Solving questions with the help of Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 4 Exploring Algebraic Identities NCERT Solutions improves confidence.
Ganita Manjari Class 9 Chapter 4 Solutions Exploring Algebraic Identities
Class 9 Ganita Manjari Chapter 4 Solutions
Class 9 Maths Ganita Manjari Chapter 4 Solutions Exploring Algebraic Identities
Think and Reflect (NCERT Textbook Page No. 69)
Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.
Solution:
Using four consecutive squares,
i.e., n2, (n + 1)2, (n + 2)2 and (n + 3)2, we find a pattern.
Here, n2 + (n + 3)2 = (n + 1)2 + (n + 2)2 + 4
For example, when n = 1
⇒ 1 + 16 = 4 + 9 + 4
⇒ 17 = 17
Think and Reflect (NCERT Textbook Page No. 71)
Question 1.
What can you say about a and b if (a + b)2 < a2 + b2?
Solution:
Given that (a + b)2 < a2 + b2
We have, (a + b)2 = a2 + 2ab + b2
So, a2 + 2ab + b2 < a2 + b2
Subtracting a2 + b2 from both sides, we get 2ab < 0, which is only possible when a and ft have opposite signs.
So, a and ft must have opposite signs for (a + b)2 < a2 + b2.
Question 2.
What can you say about a and b if (a + b)2 > a2 + b2?
Solution:
Here (a + b)2 > a2 + b2
We know (a + b)2 = a2 + 2ab + b2
∴ a2 + 2ab + b2 > a2 + b2
⇒ a2 + 2ab + b2 – a2 – b2 > a2 + b2 – a2 – b2 [Subtracting (a2 + b2) on both sides]
⇒ 2ab > 0
It is possible only when both a and b have the same sign.
Hence, a and b must have the same sign for (a + b)2 > a2 + b2.
Question 3.
When will (a + b)2 be equal to a2 + b2?
Solution:
Given that (a + b)2 = a2 + b2
We have, (a + b)2 = a2 + b2 + 2ab
So, a2 + 2ab + b2 = a2 + b2
Subtracting a2 + b2 from both sides, we get 2ab = 0, which is only possible when either a = 0 or b = 0.
So, either a = 0 or b = 0 for (a + b)2 = a2 + b2.
Did you observe that (a + b)2 and a2 + b2 are both positive? What term will decide which is larger? Use the expansion of (a + b)2 to decide.
Solution:
Yes, both (a + b)2 and a2 + b2 are indeed positive, as they are squares of numbers.
The term 2ab will determine which of the two expressions is larger. If a and b have the same sign, then 2ab > 0, so (a + b)2 > a2 + b2. If they have opposite signs, then 2 ab < 0, so (a + b)2 < a2 + b2.
Think and Reflect (NCERT Textbook Page No. 73)
What if we replace b by -b in (a + b)2 = a2 + 2ab + b2
Solution:
We get (a – b)2 = a2 – 2ab + b2, which is also an identity and can be used in ways similar to (a + b)2 = a2 + 2ab + b2
Think and Reflect (NCERT Textbook Page No. 76)
Label the squares and rectangles in Fig. 4.4 so that it represents the identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
Solution:
Think and Reflect (NCERT Textbook Page No. 78)
Question 1.
Try to evaluate the following using a suitable identity:
(i) 352
(ii) 652
(iii) 852
(iv) 1052
Do you observe any interesting pattern?
Solution:
(i) We have, a2 = (a + b)(a – b) + b2
352 = (35 + 5)(35 – 5) + 52
= 40 × 30 + 25
= 1225
(ii) 652 = (65 + 5 × 65 – 5) + 52
= 70 × 60 + 25
= 4200 + 25
= 4225
(iii) 852 = (85 + 5)(85 – 5) + 52
= 90 × 80 + 25
= 7200 + 25
= 7225
(iv) 1052= (105 + 5)(105 – 5)+ 52
= 110 × 100 + 52
= 11000 + 25
= 11025
Here we observe that to square a number ending in 5, multiply the leading part (the part except 5) by its successor and write 25 at the end.
Question 2.
Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.
Solution:
The top row shows squares with sides
(a + b + c), (a + b – c), (a – b + c), (a – b – c)
So the total area of the squares in the top row is
(a + b + c)2 + (a + b – c)2 + (a – b + c)2 + (a – b – c)2.
The bottom row rearranges all those pieces into:
- one square of side 2a
- one square of side 2b
- one square of side 2c
So, their areas are (2a)2, (2b)2, (2c)2
The total area of squares in the bottom row is
= (2a)2 + (2b)2 + (2c)2
= 4a2 + 4b2 + 4c2
= 4 (a2 + b2 + c2)
So, the algebraic identity represented:
(a + b + c)2 + (a + b – c)2 + (a-b + c)2 + (a – b – c)2 = 4 (a2 + b2 + c2).
Think and Reflect (NCERT Textbook Page No. 79)
Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
Solution:
Do it Yourself
Think and Reflect (NCERT Textbook Page No. 79)
Question 1.
Figure out the product of x + 2 and x + 3 using algebra tiles.
Solution:
We want to find (x + 2)(x + 3)
Step 1: Represent each binomial with algebra tiles
Step 2: Form a rectangle (area model) with dimensions (x + 2) by (x + 3).
Step 3: Add the areas of all tiles
x2 + x + x + x + x + x + 1 + 1 + 1 + 1 + 1 + 1
= x2 + 5x + 6
Therefore, (x + 2)(x + 3) = x2 + 5x + 6
Question 2.
Lay out algebra tiles for x2 + 11x + 30 in such a way that you will see its factors.
Solution:
Layout algebra tiles for x2 + 11x + 30
Step 1: Use the tiles x2 + 11x + 30 = 1x2 + 11x + 30.
Step 2: Arrange to form a rectangle
We have arranged the tiles into a rectangle with dimensions (x + 5) by (x + 6)
Step 3: Hence, x2 + 11x + 30 = (x + 5)(x + 6)
So, the factors are (x + 5) and (x + 6).
Think and Reflect (NCERT Textbook Page No. 80)
We have seen that (x + 3)(x + 4) = x2 + 7x + 12.
Also (x + 6)(x + 7) = x2 + 13x + 42.
Generalise the pattern to get an expression for (x + a)(x + b).
Solution:
We have (x + 3)(x + 4) = x(x + 4) + 3(x + 4)
= x2 + 4x + 3x + 12
= x2 + (3 + 4)x + 12
= x2 + 7x + 12
and (x + 6)(x + 7) = x(x + 7) + 6(x + 7)
= x2 + 7x + 6x + 42
= x2 + (7 + 6)x + 42
= x2 + 13x + 42
Hence, the general pattern for (x + a)(x + b) is x2 + (a + b)x + ab.
Think and Reflect (NCERT Textbook Page No. 82)
James and Reshma were talking about algebraic identities they learnt in school.
James: (a – b)2 (a + b) = (a2 – 2ab + b2)(a + b)
Reshma: I have a different idea: (a – b)2 (a + b) = (a – b) [(a-b) (a + b)] = (a-b)(a2 – b2)
I will find this product to get the answer.
According to you, who is correct and why?
Try to combine more such identities and find new results.
Solution:
Both are correct. They used different algebraic identities to reach the same result.
Both identities are valid, and combining them in different ways still leads to the same final polynomial (a3 – a2b – ab2 + b3).
Think and Reflect (NCERT Textbook Page No. 85)
We already know that x2 – y2 = (x – y)(x + y)
Further, we have verified that x3 – y3 = (x -y)(x2 + xy + y2) Observe that x – y is a common factor of x2 – y2 and x3 – y3. Do you think x-y is also a factor of x4 – y4?
Note that x4 – y4 = (x2)2 – (y2)2 = (x2 – y2) (x2 + y2).
Can you see how x – y is a factor of x4 – y4?
How about x5 – y5? Does this also have x – y as a factor?
Solution:
Here x4 – y4 = (x2)2 – (y2)2 = (x2 – y2) (x2 + y2)
We know that x2 – y2 = (x – y) (x + y)
So, x4 – y4 = (x – y) (x + y) (x2 + y2)
∴ Yes, (x – y) is also a factor of x4 – y4.
Now x3 – y3 = (x – y) (x2 + xy + y2)
x4 – y4 = (x – y) (x + y) (x2 + y2)
= (x – y) x(x2 + y2) + y(x2 + y2)
– (x – y) (x3 + xy2 + x2y + y3)
Now, x5 – y5 = (x – y) (x4 + x3y + xy3 + x2y2 + y4)
∴ Yes, x – y is also a factor of x5 – y5.
Think and Reflect (NCERT Textbook Page No. 87)
Try to simplify the following rational expression:
\(\frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}=\frac{(6 s-t)^2}{(?+?)(?+?)}\).
(Hint: Factor t2 + 2ts – 48s2 and simplify the rational expressions assuming that t2 + 2ts – 48s2 * 0).
Solution:
36s2 – 12st + t2 = (6s)2 – 2(6s)t + t2
= (6s – t)2
= t2 + 2ts – 48s2
= t2 + [8s + (-6s)]t + (-6s)(8s)
= [t + 8s] [t + (-6s)]
= (t + 8s) (t – 6s)
= -(t + 8s) (6s – t)
\(\frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}=\frac{(6 s-t)^2}{(t+\underline{8 s})(\underline{-6 s}+\underline{t})}=\frac{(6 s-t)(6 s-t)}{-(t+8 s)(6 s-t)}\)
= \(\frac{-(6 s-t)}{(t+8 s)}=\frac{t+6 s}{t+8 s}\)
Thus \(\frac{36 s^2-12 s t+t^2}{t^2+2 t s-48 s^2}=\frac{t+6 s}{t+8 s}\)
Ex 4.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.1 Solutions
Exercise 4.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.1 Solutions
Question 1.
Using the identity (a + b)2 = a2 + 2ab + b2, expand the following:
(i) (7x + 4y)2
Solution:
(7x + 4y)2 = (7x)2 + 2 × 7x × 4y + (4y)2
[∵ (a + b)2 = a2 + 2ab + b2] = 49x2 + 56xy + 16y2
(ii) (\(\frac{7}{5}\)x + \(\frac{3}{2}\)y)2
Solution:
(iii) (2.5p + 1.5q)2
Solution:
(2.5p + 1.5q)2
= (2.5p)2 + 2 × (2.5p) × (1.5q) + (1.5q)2
= 6.25p2 + 1.5pq + 2.25 q2
(iv) (\(\frac{3}{4}\)s + 8t)2
Solution:
(\(\frac{3}{4}\)s + 8t)2
=(\(\frac{3}{4}\)s)2 + 2(\(\frac{3}{4}\)s)(8t) + (8t)2
= \(\frac{9}{16}\)s2 + 12st + 64t2
(v) (x + \(\frac{1}{2y}\))2
Solution:
(x + \(\frac{1}{2y}\))2
= x2 + 2(x)(\(\frac{1}{2y}\)) + (\(\frac{1}{2y}\))2
= x2 + \(\frac{x}{2y}\) + \(\frac{1}{4 y^2}\)
(vi) \(\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
Solution:
\(\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
= \(\left(\frac{1}{x}\right)^2+2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)+\left(\frac{1}{y}\right)^2\)
= \(\frac{1}{x^2}+\frac{2}{x y}+\frac{1}{y^2}\)
Question 2.
Using the same identity, find the values of the following:
(i) (64)2
Solution:
(64)2
Using the identity: (a + b)2 = a2 + 2ab + b2
642 = (60 + 4)2
= 602 + 2 × 60 × 4 + 42
= 3600 + 480 + 16
= 4096
(ii) (105)2
Solution:
(105)2
Using the identity: (a + b)2 = a2 + 2ab + b2
1052 = (100 + 5)2
= 1002 + 2 × 100 × 5 + 52
= 10000+ 1000 + 25
= 11025
(iii) (205)2
Solution:
(205)2
Using the identity: (a + b)2 = a2 + 2ab + b2
2052 = (200 + 5)2
= 2002 + 2 × 200 × 5 + 52
= 40000 + 2000 + 25
= 42025
Ex 4.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.2 Solutions
Exercise 4.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.2 Solutions
Question 1.
Factor completely:
(i) 9x2 + 14xy + 16y2
Solution:
9x2 + 14xy + 16y2
= (3x)2 + 2 × (3x) × (4y) + (4y)2
= (3x + 4y)2 [∵ (a + b)2 = a2 + 2ab + b2]
(ii) 4s2 + 20st + 25t2
Solution:
4s2 + 20st + 25t2
= (2s)2 + 2 × (2s) × (5t) + (5t)2
= (2s + 5t)2 [∵ (a + b)2 = a2 + 2ab + b2]
(iii) 49x2 + 28xy + 4y2
Solution:
49x2 + 28xy + 4y2
= (7x)2 + 2 × (7x) × (2y) + (2y)2
= (7x + 2y)2 [∵ (a + b)2 = a2 + 2ab + b2]
(iv) 64p2 + \(\frac{32}{3}\)pq + \(\frac{4}{9}\)q2
Solution:
64p2 + \(\frac{32}{3}\)pq + \(\frac{4}{9}\)q2
= (8p)2 + 2 × (8p) × (\(\frac{2}{3}\)q) + (\(\frac{2}{3}\)q)2
= (8p + \(\frac{2}{3}\)q)2
(v) 3a2 + 4ab + \(\frac{4}{3}\)b2
Solution:
3a2 + 4ab + \(\frac{4}{3}\)b2
= (√3a)2 + 2 × (√3a) × (\(\frac{2}{√3}\)b) + (\(\frac{2}{√3}\)b)2
[∵ (√3) = √3 × √3 = \(\sqrt{3 \times 3}\) = 3]
= (√3a + \(\frac{2}{√3}\)b)2 [∵ (a + b)2 = a2 + 2ab + b2]
(vi) \(\frac{9}{5}\)s2 + 6sv + 5v2
Solution:
= (\(\frac{3}{\sqrt{5}}\)s)2 + 2 × (\(\frac{3}{\sqrt{5}}\)s) × (√5v) + (√5v)2
(∵ 6 = 2 × \(\frac{3}{\sqrt{5}}\) × √5)
= (\(\frac{3}{\sqrt{5}}\)s + √5v)2
[∵ (a + b)2 = a2 + 2ab + b2]
Question 2.
Find the values of the following using the identity (a – b)2 = a2 – 2ab + b2.
(i) (79)2
(ii) (193)2
(iii) (299)2
Solution:
(i) We have, (79)2 = (80 – 1)2
= (80)2 + (1)2 – 2(80)(1)
[∵ (a – b)2 = a2 + b2 – 2ab]
= 6400 + 1 – 160
= 6401 – 160
= 6241
(ii) We have, (193)2 =(200 – 7)2
= (200)2 + (7)2 – 2(200) (7)
[∵ (a -b)2 = a2 + b2 – 2ab]
= 40000 + 49 – 2800
= 40049 – 2800
= 37249
(iii) We have (299)2 =(300 – 1)2
= (300)2 + (1)2 – 2(300) (1)
[∵ (a – b)2 = a2 + b2 – 2ab]
= 90000 + 1 – 600
= 90001 – 600
= 89401
Ex 4.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.3 Solutions
Exercise 4.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.3 Solutions
Question 1.
Find the following squares using one of the above calculations easier.
(i) 1172
Solution:
1172 = (100 + 10 + 7)2
= 1002 + 102 + 72 + 2(100)(10) + 2(10)(7) + 2(7)(100)
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2be + 2ca]
= 10000 +100 + 49 + 2000 + 140 + 1400 = 13689
(ii) 782
Solution:
782 = (80 – 2)2 = 802 – 2 × 80 × 2 + 22
[∵(a – b)2 = a2 – 2ab + b2]
= 6400 – 320 + 4
= 6084
(iii) 1982
Solution:
1982 = (200 – 2)2
= 2002 + 22 – 2 × 200 × 2
[∵(a – b)2 = a2 – 2ab + b2]
= 40000 – 800 + 4
= 39204
(iv) 2142
Solution:
2142 = (200 + 10 + 4)2
= 2002 + 102 + 42 + 2(200)(10) + 2(10)(4) + 2(4)(200)
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2be + 2ca]
= 40000 + 100 + 16 + 4000 + 80 + 1600
= 45796
(v) 11042
Solution:
11042 = (1000+ 100 + 4)2
= 10002 + 1002 + 42 + 2(1000) (100) + 2(100)(4) + 2(4)(1000)
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2be + 2ca]
= 1000000 + 10000 + 16 + 200000 + 800 + 8000
= 1218816
(vi) 11202
Solution:
11202 = (1000 + 100 + 20)2
= 10002 + 1002 + 202 + 2(1000)(100) + 2(100)(20) + 2(20)(1000)
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 1000000 + 10000 + 400 + 200000 + 4000 + 40000
= 1254400
Question 2.
Factor using suitable identities:
(i) 16y2 – 24y + 9
Solution:
16y2 – 24y + 9 = (4y)2 – 2 × (4y) × 3 + (3)2
= (4y – 3)2
[∵(a – b)2 = a2 – 2ab + b2]
(ii) \(\frac{9}{4}\)s2 + 6st + 4t
Solution:
\(\frac{9}{4}\)s2 + 6st + 4t
= (\(\frac{3}{2}\)s)2 + 2 × (\(\frac{3}{2}\)s) × (2t) + (2t)2
= (\(\frac{3}{2}\)s + 2t)2
(iii) \(\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}\) + 3nk + 2mn + 9n2
Solution:
= \(\left(\frac{m}{3}\right)^2+\left(\frac{k}{2}\right)^2\) + (3n)2 + 2 × (\(\frac{m}{3}\)) × (3n) + 2 × \(\left(\frac{k}{2}\right) \times\left(\frac{m}{3}\right)\) + 2 × (\(\frac{k}{2}\)) × (3n)
= \(\left(\frac{m}{3}+\frac{k}{2}+3 n\right)^2\)
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
(iv) \(\frac{p^2}{16}\) – 2\(\frac{16}{p^2}\)
Solution:
\(\left(\frac{p}{4}\right)^2\) – 2 × \(\frac{p}{4} \times \frac{4}{p}+\left(\frac{4}{p}\right)^2\)
= \(\left(\frac{p}{4}-\frac{4}{p}\right)^2\)
[∵(a – b)2 = a2 – 2ab + b2]
(v) 9a2 + 4b2 + c2 – 12ab + 6ac – 46c
Solution:
9a2 + 4b2 + c2 – 12ab + 6ac – 46c
= (3a)2 + (-2b)2 + (c)2 + 2(3a)(-2b) + 2(3a)(c) + 2(-2b)(c)
= [3a + (-2b) + c]2
= (3a – 2b + c)2
Question 3.
Expand the following using the identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca:
(i) (p + 3q + 7r)2
Solution:
(p + 3q + 7r)2
= p2 + (3q)2 + (7r)2 + 2{p)(3q) + 2(3q)(7r) + 2(p)(7r)
= p2 + 9q2 + 49r2 + 6pq + 42 qr + 14pr
(ii) (3x – 2y + 4z)2
Solution:
(3x – 2y + 4z)2
= (3x)2 + (-2y)2 + (4z)2 + 2(3x)(-2y) + 2(-2y) + 2(-2y)(4z) + 2(3x)(4z)
= 9x2 + 4y2 + 16x2 – 12xy – 16yz + 24xz
Question 4.
Is this an identity?
(a + b – c)2 + (a – b + c)2 + (a – b – c)2 = 2a2 + 2b2 + 2c2.
Solution:
No, this is not an identity.
Expanding the LHS of the equation, we get
a2 + b2 + c2 + 2ab – 2ac – 2bc + a2 + b2 + c2 – 2ab + 2ac – 2bc + a2 + b2 + c2 – 2ab – 2ac + 2bc = 3a2 + 3b2 + 3c2 – 2bc – 2ab – 2 ac which is not equal to RHS.
Hence, the given equation is not an identity.
Ex 4.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.4 Solutions
Exercise 4.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.4 Solutions
Question 1.
Fill in the blanks to complete the following identities:
(i) s2 – 11s + 24 = (__________)(__________)
Solution:
s2 – 11s + 24
Comparing it with s2 + (a + b)s + ab, we get a + b = -11 and ab = 24.
Clearly, these two equations can be satisfied together only when a = -8 and b = -3, or vice versa.
So, s2 + [(-8) + (-3 )]s + 24 = [s + (-8)][s + (-3)]
= (s – 8)(s – 3)
[x2 + (a + b)x + ab = (x + a)(x + b)]
Thus, s2 – 11x + 24 = (s – 8) (s – 3)
(ii) (__________) (x + 1) = (3x2 – 4x – 7)
Solution:
(__________) (x + 1) = (3x2 – 4x – 7)
We have, 3x2 – 4x -7
Multiply the coefficient of x2 (which is 3) by the constant term (which is -7): 3 x (-7) = -21
Now, we need to find two numbers that multiply to -21 and add up to the coefficient of x (which is -4).
The two numbers that satisfy this condition are -7 and 3 because (-7) x 3 = —21,
(-7) + 3 = -4
Therefore, (3x2 – 4x -7) = 3x2 – (7 – 3)x – 7
= 3x2 + 3x – 7x – 7
= 3x(x + 1) – 7(x + 1)
= (3x – 7) (x + 1)
Thus, (3x – 7) (x + 1) = (3x2 – 4x – 7)
(iii) 10x2 – 11x – 6 = (2x – __) (__ + 2)
Solution:
10x2 – 11x – 6 = (2x – __) (__ + 2)
We have, 10x2 – 11x – 6
Multiply the coefficient of x2 (which is 10) by the constant term (which is -6): 10 x (-6) = -60.
Now, we need to find two numbers that multiply to -60 and add up to the coefficient of x (which is -11).
The two numbers that satisfy this condition are -15 and 4 because (-15) x 4 = -60,
(-15)+ 4 = -11.
Therefore, 10x2 – 11x – 6 = 10x2 + [4 + (-15)]x – 6
= 10x2 + 4x — 15x – 6
= 2x(5x + 2) – 3(5x + 2)
= (2x – 3)(5x + 2)
Thus, 10x2 – 11x – 6 = (2x – 3) (5x + 2)
(iv) 6x2 + 7x + 2 = (__________)(__________)
Solution:
6x2 + 7x + 2 = (__________)(__________)
Multiply the coefficient of x2 (which is 6) by the constant term (which is 2): 6 × 2 = 12.
Now, we need to find two numbers that multiply to 12 and add up to the coefficient of x (which is 7).
The two numbers that satisfy this condition are 3 and 4 because 3 × 4 = 12 and 3 + 4 = 7.
Therefore, 6x2 + 7x + 2 = 6x2 + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2)(2x + 1).
Thus, 6x2 + 7x + 2 = (3x + 2)(2x + 1)
Question 2.
Select and use the identity that will help you to find the following products without multiplying directly:
(î) (41)2
Solution:
(41)2 = (41 + 1)(41 – 1)+ 12
=42 × 40 + 1
= 1681
[∵ a2 = (a + b)(a – b) + b2]
(ii) (27)2
Solution:
(27)2 = (30 – 3)2 = 3022 × 30 × 3 + 32
=900 – 180 + 9
=729
[∵ (a – b)2 = a2 – 2ab + b2]
(iii) (23 × 17)
Solution:
(23 × 17) = (20 + 3) (20 – 3) = 202 – 32
= 400 – 9
= 391
[∵ a2 – b2 = (a + b)(a – b)]
(iv) (135)2
Solution:
(135)2 = (140 – 5)2
= 1402 – 2 × 140 × 5 + 52
= 19600 – 1400 + 25
= 18225
[∵ (a – b)2 = a2 – 2ab + b2]
(v) (97)2
Solution:
(97)2 = (97 + 7) (97 – 7) + 72
= 104 × 90 + 49
= 9409
[∵ a2 = (a + b)(a – b) + b2]
(vi) (18 × 29)
Solution:
(18 × 29) = (20 – 2) (20 + 9)
= 202 + (-2 + 9)20 + (-2) × 9
[∵ (x + a)(x + b) = x2 + (a + b)x + ab]
= 400 + 140 – 18
= 522
(vii) (34 × 43)
Solution:
= (40 – 6) (40 + 3)
= 402 + (-6 + 3)40 + (-6) × 3
[∵ (x + a)(x + b) = x2 + (a + b)x + ab]
= 402 + (-6 + 3)40 + (-6) × 3
= 1600 – 120 – 18
= 1462
(viii) (205)2
Solution:
(205)2 = (205 + 5) (205 – 5) + 52
= 210 × 200 + 25
= 42025
[∵ a2 = (a + b)(a – b) + b2]
Question 3.
Factor the following:
(i) 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc
Solution:
9a2 + b2 + 4c2 – 6ab + 12ac – 4be
= (3a)2 + (-b)2 + (2c)2 + 2(3a) (-b) + 2(3a)(2c) + 2(-b)(2c)
= (3a – b + 2c)2
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
(ii) 16s2 + 25t2 – 40st?
Solution:
16s2 + 25t2 – 40st
= (4s)2 + (5t)2 – 2 × (4s)(5t)
= (4s – 5t)2
[∵ (a – b)2 = a2 + b2 – 2ab]
(iii) r2 – r – 42
Solution:
r2 – r – 42
= r2 + (-7 + 6)r + (—7)(6)
= [r+ (-7)] (r + 6)
[∵ (x + a)(x + b) = x2 + (a + b)x + ab]
= (r – 7)(r + 6)
(iv) 49g2 + 14gb + b2
Solution:
49g2 + 14gb + b2
= (7g)2 + 2(7g)b + h2
= (7g + h)2 [∵ (a + b)2 = a2 + b2 + 2ab]
(v) 64u2 + 121 v2 + 4W2 – 176uv – 32uw + 44vw
Solution:
64u2 + 121 v2 + 4W2 – 176uv- 32uw + 44vw
= (-8u)2 + (11 v)2 + (2w)2 + 2(-8u)(11 v) + 2(-8u)(2v) + 2(11v) (2w)
= [(-8)u + 11v + 2w]2
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= (11v + 2w – 8u)2
Ex 4.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.5 Solutions
Exercise 4.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.5 Solutions
Question 1.
Simplify the following rational expressions, assuming that the expressions in the denominators are not equal to zero:
(i) \(\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}\)
Solution:
\(\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}\)
We have,
3p2 – 3pq – 18q2 = 3 (p2 – pq – 6q2)
= 3(p2 – 3pq + 2pq – 6q2)
(∵ -6 = -3 × 2 and -3 + 2 = -1)
= 3[p(p – 3q) + 2q(p – 3q)]
3p2 – 3pq – 18q2 = 3(p + 2q)(p – 3q) …(i)
Now, p2 + 3pq – 10q2 = p2 + 5pq – 2pq – 10q2
[∵ -10 = 5 × (-2) and 5 + (-2) = 3]
= p(p + 5q) – 2q(p + 5q)
Therefore, p2 + 3pq – 10q2 = (p – 2q)(p + 5q) …(ii)
From (i) and (ii), we get
\(\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}=\frac{3(p+2 q)(p-3 q)}{(p-2 q)(p+5 q)}\)
(ii) \(\frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2}\)
Solution:
Given the expression: \(\frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2}\)
Step 1: Factorise the numerator and denominator.
Numerator: The expression n3 – 3n2m + 3nm2 – m3 is a perfect cube expansion and factors as (n – mf
Denominator: The expression 5m2 – 10mn + 5n2 factors as 5(m – n)2
Step 2: Simplify the expression.
Now, the expression becomes \(\frac{(n-m)^3}{5(m-n)^2}\)
Since (n – m) = -(m – n), we get
\(\frac{-(m-n)^3}{5(m-n)^2}=\frac{-(m-n)}{5}=\frac{n-m}{5}\)
(iii) \(\frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}\)
Solution:
We have,
\(\frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}\)
Using the identity a3 + b3 + c3 – 3abc = (a + b + c)
(a2 + b2 + c2 -ab – bc – ca),
take a = w,b = -v, c = x, then
w3 – v3 + x3 + 3wvx = (w – v + x) (w2 + v2 + x2 + wv + vx – wx)
Also, the denominator
w2 + v2 + x2 – 2wv – 2vx + 2wx = (w – v + x)2
Hence,
= \(\frac{(w-v+x)\left(w^2+v^2+x^2+w v+v x-w x\right.}{(w-v+x)^2}\)
= \(\frac{w^2+v^2+x^2+w v+v x-w x}{w-v+x}\)
(iv) \(\frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)}\)
Solution:
\(\frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)}\)
4y2 – 20yz + 25z2 = (2y – 5z)2
25z2 – 4y2 = (5z – 2y)(5z + 2y) = \(\frac{(2 y-5 z)^2}{(5 z-2 y)(5 z+2 y)}\)
Since (2y – 5z) = -(5z – 2y), (2y – 5z)2 = (5z – 2y)2
So, \(\frac{(5 z-2 y)^2}{(5 z-2 y)(5 z+2 y)}=\frac{5 z-2 y}{5 z+2 y}\)
(v) \(\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}\)
Solution:
We have,
\(\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}\)
Now, numerator = (x2 + x – 6)(x2 -7x + 12)
= [x2 + 3x – 2x – 6].[x2 – 4x – 3x + 12]
= [x(x + 3) – 2(x + 3)].[x(x – 4) -3(x – 4)]
= [(x + 3)(x – 2)].[(x – 4)(x – 3)]
and denominator = (x2 – 6x + 8)(x2 – 9)
= [x2 – 4x – 2x + 8].(x – 3)(x + 3)
[va2 – b2 = (a + b)(a – b)]
= [x(x – 4) – 2(x – 4)] .(x – 3)(x + 3)
= (x – 4) (x – 2) (x – 3) (x + 3)
Now,
\(\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}\)
= \(\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}\) = 1
(vi) \(\frac{p^4-16}{p^2-4 p+4}\)
Solution:
Ganita Manjari Class 9 Maths Chapter 4 End of Chapter Exercise Solutions
Exploring Algebraic Identities End of Chapter Exercise Solutions
Question 1.
Use suitable identities to find the following products:
(i) (-3x + 4)2
Solution:
(-3x + 4)2
We can expand this using the identity
(a + b)2 = a2 + 2ab + b2
(-3x + 4)2 = (-3x)2 + 2(-3x)(4) + 42
= 9x2 – 24x + 16
(ii) (2s + 7) (2s – 7)
Solution:
(2s + 7) (2s – 7)
We can use the identity
(a + b)(a – b) = a2 – b2
(25 + 7) (25 – 7) = (25)2 – (7)2
= 4s2 – 49
(iii) (p2 + \(\frac{1}{2}\))(p2 – \(\frac{1}{2}\))
Solution:
(p2 + \(\frac{1}{2}\))(p2 – \(\frac{1}{2}\))
Using the identity: (a + b)(a – b) = a2 – b2
(p2 + \(\frac{1}{2}\))(p2 – \(\frac{1}{2}\))
= (p2)2 – (\(\frac{1}{2}\))2
= p4 – \(\frac{1}{4}\)
(iv) (2n + 7) (2n – 7)
Solution:
(2n + 1)(2n – 1)
Using the identity: (a + b)(a – b) = a2 – b2
(2n + 7) (2n – 7) = (2n)2 – 72
= 4n2 – 49
(v) (s – 2t) (s2 + 2st + 4t2)
Solution:
(s – 2t) (s2 + 2st + 4t2)
Here, we use the identity
(x – y)(x2 + xy + y2) = x3 – y3
(s – 2t) (s2 + 2st + 4t2) = s2 – (2t)2
= s3 – 8t3
(vi) (\(\frac{1}{2r}\) – 4r)2
Solution:
(\(\frac{1}{2r}\) – 4r)2
We simplify this by using the identity
(a – b)2 = a2 – 2ab + b2
(\(\frac{1}{2r}\))2 – 2(\(\frac{1}{2r}\))(4r) + (4r)2
= \(\frac{1}{4 r^2}\) – 4 + 16r2
(vii) (-3m + 4k – l)2
Solution:
(-3m + 4k – l)2
We can expand using the identity:
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
(-3m + 4k – l)2 = (—3m)2 + (4k)2 + (-l)2 + 2(-3m)(4k) + 2(-3m)(-l) + 2(4k)(-l)
= 9m2 + 16k2 +12 — 24mk + 6ml – 8kl
(viii) (x – \(\frac{1}{3}\))3
Solution:
(x – \(\frac{1}{3}\))3
To expand this, we use the identity:
(a – b)3 = a3 – 3a2b + 3ab2 – b3
= a3 – b3 – 3ab(a – b)
(x – \(\frac{1}{3}\))3 = x3 – \(\frac{1}{27}\)y3 – 3x x \(\frac{1}{3}\)y(x – \(\frac{1}{3y}\))
= x3 – \(\frac{1}{27}\)y – x2y + \(\frac{x}{3}\)
(ix) (\(\frac{7}{2}\)k – \(\frac{2}{3}\)m)3
Solution:
To expand this, we use the identity:
(a – b)3 = a3 – 3a2b + 3ab2 – b3
Question 2.
Find the values using suitable identities:
(i) 17 × 21
Solution:
17 × 21
Here, 17 = 20 – 3 and 21 = 20 + 1, so:
17 × 21 = (20 – 3)(20 + 1)
Using the identity (x + a)(x + b)
= x2 + (a + h)x + ab
= 202 + (-3 + 1)20 + (-3)(1)
= 400 – 40 – 3
= 357
(ii) 104 × 96
Solution:
104 × 96
Here, 104 = 100 + 4 and 96 = 100 – 4,
So, 104 × 96 = (100 + 4)(100 – 4)
Using the identity (a + b)(a – b) = a2 – b2
= 1002 – 42
= 10000 – 16
= 9984
(iii) 24 × 16
Solution:
24 × 16
Here, 24 = 20 + 4 and 16 = 20 – 4
So, 24 × 16 = (20 + 4)(20 – 4)
Using the identity (a + b)(a – b) = a2 – b2
= 202 – 42
= 400- 16
= 384
(iv) 1473
Solution:
1473
We can use the identity for cubes:
(a – b)3 = a3 – 3a2b + 3 ab2 – b3
Here, 147 = 150 – 3,
So, 1473 = (150 – 3)3
= 1503 – 3 × 1502 × 3 + 3 × 150 × 32 – 33
Simplifying:
= 3375000 – 202500 + 4050 – 27
= 3176523
(v) 1993
Solution:
1993
We can use the identity for cubes:
(a – b)3= a3 – 3a2b + 3ab2 – b3
Here, 199 = 200- 1,
So, 1993 = (200 – 1)3
= 2003 – 3 × 2002 × 1 + 3 × 200 × 12 – 13
= 8000000 – 120000 + 600 – 1
= 7880599
(vi) 1273
Solution:
1273
We can use the identity for cubes:
(a – b)3 = a3 – 3a2b + 3ab2 – b3
Here, 127 = 130 – 3,
So, 1273 = (130 – 3)3
= 1303 – 3 × 1302 × 3 + 3 × 130 × 32 – 33
= 2197000 – 152100 + 3510 – 27
= 2048383
(vii) (-107)3
Solution:
(-107)3
We can use the identity for cubes:
(a – b)3 = a3 – 3a2b + 3ab2 – b3
Here, -107 =-100 – 7,
So, (-107)3 = (-100 – 7)3
= (-100)3 – 3 × (-100)2 × 7 + 3 × (-100) × 72 – 73
= -1000000 – 210000 + (-14700) – 343
= -1225043
(viii) (-299)3
Solution:
(-299)3
We can use the identity for cubes:
(a + b)3 = a3 + 3a2b + 3 ab2 + b3
Here, -299 = -300 + 1,
So, (-299)3 = (-300 + 1)3
= (-300)3 + 3 × (-300)2 × 1 + 3 × (-300) × 12+ 13
= -27000000 + 270000 – 900 + 1
= 26730899
Question 3.
Factor the following algebraic expressions:
(i) 4y2 + 1 + \(\frac{1}{16 y^2}\)
Solution:
4y2 + 1 + \(\frac{1}{16 y^2}\)
= (2y)2 + 2 × (2y) × \(\left(\frac{1}{4 y}\right)+\left(\frac{1}{4 y}\right)^2\)
= (2y + \(\frac{1}{2y}\))2
[∵ (a + b)2 = a2 + 2ab + b2]
(ii) 9m2 – \(\frac{1}{25 n^2}\)
Solution:
9m2 – \(\frac{1}{25 n^2}\)
= (3m)2 – (\(\frac{1}{5n}\))2
= (3m + \(\frac{1}{5n}\))(3m – \(\frac{1}{5n}\))
[∵ a2 – b2 = (a + b)(a – b)]
(iii) 27b3 – \(\frac{1}{64 b^3}\)
Solution:
(iv) x2 + \(\frac{5 x}{6}+\frac{1}{6}\)
Solution:
(v) 27u3 – \(\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25}\)
Solution:
(vi) 64y3 – \(\frac{1}{125}\)z3
Solution:
(vii) p3 + 27q3 + r3 – 9pqr
Solution:
p3 + 27q3 + r3 – 9pqr
We have, x3 + y3 + z3 – 3xyz
= (x + y + z)(x3 + y3 + z3 – xy – xz – yz)
= (p + 3q + r)(p3 + 9q3 + r3 – 3pq – 3qr – pr)
(viii) 9m2 – 12m + 4
Solution:
9m2 – 12m + 4
= (3m)2 – 2(3m)(2) + (2)2
= (3m – 2)2
[∵ (a – b)2 = a2 – 2ab + b2]
(ix) 9x3 – \(\frac{8}{3}\) y3 + \(\frac{z^3}{3}\) + 6xyz
Solution:
9x3 – \(\frac{8}{3}\) y3 + \(\frac{z^3}{3}\) + 6xyz
= \(\frac{1}{3}\)(27x3 – 8y3 + z3 + 18xyz)
= \(\frac{1}{3}\)[(3x)3 + (-2y)3 + (z)3 – 3(3x)(-2y)(z)]
= \(\frac{1}{3}\)(3x – 2y +z)(9x3 + 4y3 + z3 + 6xy + 2yz – 3xz)
[∵ a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
(x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy
Solution:
4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy
= (2x)2 + (3y)2 + (6z)2 + 2(2x)(3y) + 1(3y)(6z) + 2(2x)(6z)
= (2x + 3y + 6z)2
[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca)
(xi) 27u3 – \(\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4}\)
Solution:
27u3 – \(\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4}\)
= (3u)3 – (\(\frac{1}{6}\))3 – 3(3u)2(\(\frac{1}{6}\)) + 3(3u)(\(\frac{1}{6}\))2
= (3u – \(\frac{1}{6}\))3
[∵ (a – b)3 = a3 – b3 – 3a2b + 3ab2]
Question 4.
Simplify the following:
(i) \(\frac{4 x^2+4 x+1}{4 x^2-1}\)
(ii) \(\frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2}\)
(iii) \(\frac{s^3+125 t^3}{s^2-2 s t-35 t^2}\)
Note: Assume that the denominators are not equal to 0.
Solution:
Question 5.
Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) 25a2 – 30ab + 9b2
Solution:
25a2 – 30ab + 9b2
= (5a)2 – 2(5a)(3b) + (3b)2
= (5a – 3b)2
[∵ (a – b) = a2 – 2ab + b2]
= (5a – 3b)(5a – 3b)
Area of a rectangle = length × breadth
Possible dimensions:
Length = (5a – 3b) units
Breadth = (5a – 3b) units
(ii) 36s2 – 49t2
Solution:
36s2 – 49t2
= (6s)2 – (7t)2
= (6s + 7t)(6s – 7t)
[∵ a2 – b2 = (a + b) (a – b)]
Area of a rectangle = length × breadth
Possible dimensions:
Length = (6s + 7t) units
Breadth = (6s – 7t) units
Question 6.
Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) 6a2 – 24b2
Solution:
6a2 – 24b2
= 6(a2 – 4b2)
= 6[a2 -(2b)2]
= 6(a + 2 b)(a – 2b) [∵ a2 – b2 = (a + b)(a – b)]
Volume of a cuboid = length × breadth × height
Possible dimensions:
Length = 6 units
Breadth = (a + 2b) units
Height = (a – 2b) units
(ii) 3ps2 – 15ps +12p
Solution:
3ps2 – 15ps + 12p
= 3p(s2 – 5s + 4)
= 3p(s2 – 4s – s + 4)
= 3p[s(s – 4) – 1(s – 4)]
= 3p(s – 1)(s – 4)
Volume of a cuboid = length × breadth × height
Possible dimensions:
Length = 3p units
Breadth = (s – 1) units
Height = (s – 4) units
Question 7.
The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Solution:
Area of the playground = 40 m × 40 m
= (40)2 m2
Area of the playground along with the path
= (40 + 2s)m × (40 + 2s)m
= (40 + 2s)2 m2
Area of the path = (40 + 2s)2 – (40)2
= (40 + 2s + 40)(40 + 2s-40) [v a2 – b2 = (a + b)(a – b)]
= (80 + 2s)(2s)
= 160s + 4s2
So, the area of the path is (4s2 + 160s) m2.
Question 8.
If a number plus its reciprocal equals \(\frac{10}{3}\), find the number.
Solution:
Let the number be x
According to the statement, we have
x + \(\frac{1}{x}=\frac{10}{3}\)
On multiplying both sides by lx, we get
⇒ 3x2 + 3 = 10x
⇒ 3x2 – 10x + 3 = 0
Now, splitting the middle term, we get
⇒ 3x2 – 9x – x + 3 = 0
⇒ 3x(x – 3) – 1(x – 3) = 0
⇒ (3x – 1) (x – 3) = 0
⇒ 3x – 1 = 0 or x – 3 = 0
⇒ x = \(\frac{1}{3}\) or x = 3
Hence, the required numbers are – and 3.
Question 9.
A rectangular pool has area 2a2 + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Solution:
Area of the pool = 2x2 + 7x + 3 square hastas
= 2x2 + 6x + x + 3 square hastas
= 2x(x + 3) + 10c+ 3) square hastas
= (2x + 1)(x + 3) square hastas
Area of the pool = length × breadth
Width of the pool = (2x + 1) hastas (given)
Therefore, length of the pool = (x + 3) hastas
Question 10.
If both x – 2 and x – \(\frac{1}{2}\) are factors of px2 + 5x + r, show 2 that p = r.
Solution:
px2 + 5x + r
Factors = x — 2, x – \(\frac{1}{2}\)
⇒ x = 2 or \(\frac{1}{2}\)
For x = 2,
p(2)2 + 5(2) + r = 0
4p + 10 + r = 0 …(1)
For x = \(\frac{1}{2}\)
p(\(\frac{1}{2}\))2 + 5(\(\frac{1}{2}\)) + r = 0
\(\frac{p}{4}+\frac{5}{2}\) + r = 0 …….(2)
Equating (1) and (2), we get
p = -2
Substituting p = -2 in equation (1), we get
4(-2) + 10 + r = 0
⇒ -8+ 10 + r = 0
⇒ 2 + r = 0
⇒ r = -2
∴ p = r = – 2
Hence, p = r
Question 11.
If a + b + c = 5 and ab + bc + ca – 10, then prove that a3 + b3 + c3 – 3abc = – 25.
Solution:
We know that
a3 + b3 + c3 – 3abc
= (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
= (a + b + c) [a2 + b2 + c2 – (ab + be + ca)]
= 5(a2 + b2 + c2 – (ab + bc + ca))
= 5(a2 + b2 + c2 – 10) …(1)
Now, a + b + c = 5
Squaring both sides, we get
(a + b + c)2 = 52
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 25
⇒ a2 + b2 + c2 + 2(10) = 25
⇒ a2 + b2 + c2 = 25 – 20 = 5
Substituting a2 + b2 + c2 = 5 in equation (1)
5(5 – 10) = 5(-5) = — 25
Hence proved.
Question 12.
By factoring the expression, check that n3 – n is, always divisible by 6 for all natural numbers n. Give reasons.
Solution:
Given, n3 – n
Now, factorising the expression, we get n3 – n = n(n2 – 1)
Also, using the identity a2 – b2 = (a – b) (a + b), we get
n(n2 – 1) = n(n – 1) (n + 1)
Now, n – 1, n, n + 1 are three consecutive natural numbers, and among any three consecutive numbers, one is divisible by 2, and also one is divisible by 3.
Hence, the product (n – 1) (n + 1) is divisible by 2 × 3 = 6.
Therefore, n3 – n is always divisible by 6 for all natural numbers n.
Question 13.
Find the value of
(i) x3 + y3 – 12xy + 64, when x + y = – 4
Solution:
x3 + y3 – 12xy – 64
Using the identity: (a + b)3 = a3 + b3 + 3ab(a + b)
(x + y)3 = x3 + y3 + 3xy(x + y)
⇒ (-4)3 = x3 + y3 + 3xy(—4)
[∵ x +y = —4]
⇒ -64 = x3 + y3 – 12xy
⇒ x3 + y3 — 12xy + 64 = 0
(ii) x3 – 8y3 – 36xy – 216, when x = 2y + 6
Solution:
We are given, x = 2y + 6
⇒ x – 2y = 6
Now, x3 – 8y3 – 36xy – 216
Using the identity: (a – b)3 = a3 – b3 – 3ab(a – b)
(x – 2y)3 = x3 – (2y)3 – 3x(2y)(x – 2y)
(6)3 = x3 – 8y3 – 6xy(6)
216 = x3 – 8y3 – 36xy
⇒ x3 – 8y3 – 36xy – 216 = 0
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