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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4

Question 1.
Expand the following:
(i) (2x + 3y + 4z)2
(ii) (-p + 2q + 3r)2
(iii) (2p + 3) (2p – 4) (2p – 5)
(iv) (3a + 1) (3a – 2) (3a + 4)
Solution:
We know that (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
(i) (2x + 3y + 4z)2 = (2x)2 + (3y)2 + (4z)2 + 2(2x) (3y) + 2(3y) (4z) + 2(4z) (2x)
= 4x2 + 9y2 + 16z2 + 12xy + 24yz + 16xz

(ii) (-p + 2q + 3r)2 = (-p)2 + (2q)2 + (3r)2 + 2(-p) (2q) + 2(2q)(3r) + 2(3r) (- p)
= p2+ 4q2 + 9r2 – 4pq + 12qr – 6pr (iii) (2p + 3) (2p – 4) (2p – 5)
[Here x = 2p, a = 3, b = -4 and c = -5]
= (2p)3 + (3 – 4 – 5) (2p)2 + [(3)(-4) + (-4)(-5) + (3) (-5)] 2p + (3) (-4) (-5)
= 8p3 + (-6)(4p2) + (-12 + 20 – 15) 2p + 60
= 8p3 – 24p2 – 14p + 60

(iv) (3a + 1) (3a – 2) (3a + 4)
[Here x = 3a, a = 1, b = -2 and c = 4]
= (3a)3 + (1 – 2 + 4) (3a)2 + [(1)(-2) + (-2) (4) + (4) (1)] (3a) + (1) (-2) (4)
= 27a3 + 3(9a2) + (-2 – 8 + 4) (3a) – 8
= 27a3 + 27a2 – 18a – 8 Question 2.
Using algebraic identity, find the coefficients of x2, x and constant term without actual expansion.
(i) (x + 5)(x + 6)(x + 7)
Solution:
[Here x = x, a = 5, b = 6, c = 7]
(x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ac)x + abc
coefficient of x2 = 5 + 6 + 7
= 18
coefficient of x = 30 + 42 + 35
= 107
constant term = (5) (6) (7)
= 210 (ii) (2x + 3)(2x – 5) (2x – 6)
Solution:
[Here x = 2x, a = 3, b = -5, c = -6]
(x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ac)x + abc
coefficient of x2 = (3 – 5 – 6)4 [(2x)2 = 4x2]
= (-8) (4)
= -32
coefficient of x = [(3)(-5) + (-5)(-6) + (-6)(3)](2)
= (-15 + 30-18) (2)
= (-3) (2)
= -6
constant term = (3) (-5) (-6)
= 90 Question 3.
If (x + a)(x + b)(x + c) = x3 + 14x2 + 59x + 70, find the value of
(i) a + b + c
(ii) $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$
(iii) a2 + b2 + c2
(iv) $$\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab}$$
Solution:
(x + a) (x + b) (x + c) = x3 + 14x2 + 59x + 70
x3 + (a + b + c)x2 + (ab + bc + ac)x + abc = x3 + 14x2 + 59x + 70
a + b + c = 14, ab + bc + ac = 59, abc = 70
(i) a + b + c = 14

(ii) $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$ = $$\frac{bc+ac+ab}{abc}$$
= $$\frac{59}{70}$$

(iii) a2 + b2 + c2 = (a + b + c)2 – 2 (ab + bc + ac)
= (14)2 – 2(59)
= 196 – 118
= 78  Question 4.
Expand:
(i) (3a – 4b)3
Solution:
(a – b)3 = a3 – b3 – 3ab (a – b)
(3a – 4b)3 = (3a)3 – (4b)3 – 3(3a)(4b)(3a – 4b)
= 27a3 – 64b3 – 36ab (3a – 4b)
= 27a3 – 64b3 – 108a2b + 144ab2

(ii) [x + $$\frac{1}{y}]^{3}$$
Solution:
(a + b)3 = a3 + b3 + 3ab (a + b) Question 5.
Evaluate the following by using identities:
(i) 983
Solution:
983 = (100 – 2)3 [(a – b)3 = a3 – b3 – 3ab (a – b)]
= 1003 – (2)3 – 3(100) (2) (100 – 2)
= 1000000 – 8 – 600(98)
= 1000000 – 8 – 58800
= 1000000 – 58808
= 941192

(ii) 10013
Solution:
(1001)3 = (1000 + 1)3
[(a + b)3 = a3 + b3 + 3ab (a + b)]
= (1000)3 + 13 + 3(1000) (1) (1000 + 1)
= 1000000000 + 1 + 3000 (1001)
= 1000000001 + 3003000
= 1003003001 Question 6.
If (x + y + z) = 9 and (xy + yz + zx) = 26, then find the value of x2 + y2 + z2.
Solution:
x + y + z = 9; xy + yz + zx = 26
x2 + y2 + z2 = (x + y + z)2 – 2xy – 2yz – 2xz
= (x + y + z)2 – 2 (xy + yz + zx)
= 92 – 2(26)
= 81 – 52
= 29

Question 7.
Find 27a3 + 64b3, If 3a + 4b = 10 and ab = 2
Solution:
3a + Ab = 10, ab = 2
27a3 + 64b3 = (3a)3 + (4b)3
[a3 + b3 = (a + b)3 – 3 ab (a + b)]
= (3a + 4b)3 – 3 × 3a × 4b (3a + 4b)
= 103 – 36ab (10)
= 1000 – 36(2)(10)
= 1000 – 720
= 280 Question 8.
Find x3 – y3, if x – y = 5 and xy = 14.
Solution:
x – y = 5, xy = 14
x3 – y= (x – y)3 + 3xy (x – y)
= 53 + 3(14) (5)
= 125 + 210
= 335

Question 9.
If a + $$\frac{1}{a}$$ = 6, then find the value of a3 +$$\frac{1}{a^3}$$
Solution:
a + $$\frac{1}{a}$$ = 6 [a3 + b3 = (a + b)3 – 3ab (a + b)] = 63 – 3(6)
= 216 – 18
= 198 Question 10.
If x2 + $$\frac{1}{x^2}$$ = 23, then find the value of x + $$\frac{1}{x}$$ and x3 + $$\frac{1}{x^3}$$
Solution: When x = 5 [a3 + b3 = (a + b)3 – 3ab (a + b)]
= (5)3 – 3(5)
= 125 – 15
= 110
when x = -5
x3 + $$\frac{1}{x^3}$$ = (-5)3 – 3(-5)
= -125 + 15
= -110
∴ x3 + $$\frac{1}{x^3}$$ = ±110

Question 11.
If (y – $$\frac{1}{y})^{3}$$ = 27 then find the value of y3 – $$\frac{1}{y^3}$$
Solution: = 33 + 3(3)
= 27 + 9
= 36 Question 12.
Simplify:
(i) (2a + 3b + 4c) (4a2 + 9b2 + 16c2 – 6ab – 12bc – 8ca)
(ii) (x – 2y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3xz)
Solution:
x3 + y3 + z3 – 3xyz ≡ (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
(i) (2a + 3b + 4c) (4a2 + 9b2 + 16c2 – 6ab – 12bc – 8ea)
= (2a)3 + (3b)3 + (4c)3 – 3 (2a) (3b) (4c)
= 8a3 + 27b3 + 64c3 – 72abc

(ii) (x – 2y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3xz)
= x3 + (-2y)3 + (3z)3 – 3(x) (-2y) (3z)
= x3 – 8y3 + 27z3 + 18xyz

Question 13.
By using identity evaluate the following:
(i) 73 – 103 + 33
Solution:
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
We know that a + b + c = 0 then a3 + b3 + c3 = 3ab
a + b + c = 7 + (-10) + 3
= 10 – 10
= 0
∴ 73 – 103 + 33 = 3(7) (-10) (3)
= -630

(ii) 1 + $$\frac{1}{8}$$ – $$\frac{27}{8}$$
Solution:
We know that a3 + b3 + c3 = 0 then a + b + c = 3abc  Question 14.
If 2x -3y – 4z = 0, then find 8x3 – 27y3 – 64z3.
Solution:
We know x3 +y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
x3 + y3 + z3 = (x + y + z) (x2 +y2 + z2 – xy – yz – zx) + 3xyz
8x3 – 27y3 – 64z3 = (2x)3 + (-3y)3 + (-4z)3
= (2x – 3y- 4z) [(2x)2 + (-3y)2 + (-4z)2 – (2x)(-3y) – (-3y) (-4z) -(-4z)(2x)] + 3(2x)(-3y)(-4z)
= 0 (4x2 + 9y2 + 16z2 + 6xy – 12yz + 8xz) + 72xyz
= 72xyz
8x3 – 27y3 – 64z3 = 72xyz