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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} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac

(i) (2x + 3y + 4z)^{2} = (2x)^{2} + (3y)^{2} + (4z)^{2} + 2(2x) (3y) + 2(3y) (4z) + 2(4z) (2x)

= 4x^{2} + 9y^{2} + 16z^{2} + 12xy + 24yz + 16xz

(ii) (-p + 2q + 3r)^{2} = (-p)^{2} + (2q)^{2} + (3r)^{2} + 2(-p) (2q) + 2(2q)(3r) + 2(3r) (- p)

= p^{2}+ 4q^{2} + 9r^{2} – 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)

= 8p^{3} + (-6)(4p^{2}) + (-12 + 20 – 15) 2p + 60

= 8p^{3} – 24p^{2} – 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)

= 27a^{3} + 3(9a^{2}) + (-2 – 8 + 4) (3a) – 8

= 27a^{3} + 27a^{2} – 18a – 8

Question 2.

Using algebraic identity, find the coefficients of x^{2}, 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) = x^{3} + (a + b + c)x^{2} + (ab + bc + ac)x + abc

coefficient of x^{2} = 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) = x^{3} + (a + b + c)x^{2} + (ab + bc + ac)x + abc

coefficient of x^{2} = (3 – 5 – 6)4 [(2x)^{2} = 4x^{2}]

= (-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) = x^{3} + 14x^{2} + 59x + 70, find the value of

(i) a + b + c

(ii) \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\)

(iii) a^{2} + b^{2} + c^{2}

(iv) \(\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab}\)

Solution:

(x + a) (x + b) (x + c) = x^{3} + 14x^{2} + 59x + 70

x^{3} + (a + b + c)x^{2} + (ab + bc + ac)x + abc = x^{3} + 14x^{2} + 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) a^{2} + b^{2} + c^{2} = (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} = a^{3} – b^{3} – 3ab (a – b)

(3a – 4b)^{3} = (3a)^{3} – (4b)^{3} – 3(3a)(4b)(3a – 4b)

= 27a^{3} – 64b^{3} – 36ab (3a – 4b)

= 27a^{3} – 64b^{3} – 108a^{2}b + 144ab^{2}

(ii) [x + \(\frac{1}{y}]^{3}\)

Solution:

(a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)

Question 5.

Evaluate the following by using identities:

(i) 98^{3}

Solution:

98^{3} = (100 – 2)^{3} [(a – b)^{3 }= a^{3} – b^{3} – 3ab (a – b)]

= 100^{3} – (2)^{3} – 3(100) (2) (100 – 2)

= 1000000 – 8 – 600(98)

= 1000000 – 8 – 58800

= 1000000 – 58808

= 941192

(ii) 1001^{3}

Solution:

(1001)^{3 }= (1000 + 1)^{3}

[(a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)]

= (1000)^{3} + 1^{3} + 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 x^{2} + y^{2} + z^{2}.

Solution:

x + y + z = 9; xy + yz + zx = 26

x^{2} + y^{2} + z^{2} = (x + y + z)^{2} – 2xy – 2yz – 2xz

= (x + y + z)^{2} – 2 (xy + yz + zx)

= 9^{2} – 2(26)

= 81 – 52

= 29

Question 7.

Find 27a^{3} + 64b^{3}, If 3a + 4b = 10 and ab = 2

Solution:

3a + Ab = 10, ab = 2

27a^{3} + 64b^{3} = (3a)^{3} + (4b)^{3}

[a^{3} + b^{3} = (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 x^{3} – y^{3}, if x – y = 5 and xy = 14.

Solution:

x – y = 5, xy = 14

x^{3} – y^{3 }= (x – y)^{3} + 3xy (x – y)

= 5^{3} + 3(14) (5)

= 125 + 210

= 335

Question 9.

If a + \(\frac{1}{a}\) = 6, then find the value of a^{3} +\(\frac{1}{a^3}\)

Solution:

a + \(\frac{1}{a}\) = 6 [a^{3} + b^{3} = (a + b)^{3} – 3ab (a + b)]

= 6^{3} – 3(6)

= 216 – 18

= 198

Question 10.

If x^{2} + \(\frac{1}{x^2}\) = 23, then find the value of x + \(\frac{1}{x}\) and x^{3} + \(\frac{1}{x^3}\)

Solution:

When x = 5 [a^{3} + b^{3} = (a + b)^{3} – 3ab (a + b)]

= (5)^{3} – 3(5)

= 125 – 15

= 110

when x = -5

x^{3} + \(\frac{1}{x^3}\) = (-5)^{3} – 3(-5)

= -125 + 15

= -110

∴ x^{3} + \(\frac{1}{x^3}\) = ±110

Question 11.

If (y – \(\frac{1}{y})^{3}\) = 27 then find the value of y^{3} – \(\frac{1}{y^3}\)

Solution:

= 3^{3} + 3(3)

= 27 + 9

= 36

Question 12.

Simplify:

(i) (2a + 3b + 4c) (4a^{2} + 9b^{2} + 16c^{2} – 6ab – 12bc – 8ca)

(ii) (x – 2y + 3z) (x^{2} + 4y^{2} + 9z^{2} + 2xy + 6yz – 3xz)

Solution:

x^{3} + y^{3} + z^{3} – 3xyz ≡ (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)

(i) (2a + 3b + 4c) (4a^{2} + 9b^{2} + 16c^{2} – 6ab – 12bc – 8ea)

= (2a)^{3} + (3b)^{3} + (4c)^{3} – 3 (2a) (3b) (4c)

= 8a^{3} + 27b^{3} + 64c^{3} – 72abc

(ii) (x – 2y + 3z) (x^{2} + 4y^{2} + 9z^{2} + 2xy + 6yz – 3xz)

= x^{3} + (-2y)^{3} + (3z)^{3} – 3(x) (-2y) (3z)

= x^{3} – 8y^{3} + 27z^{3} + 18xyz

Question 13.

By using identity evaluate the following:

(i) 7^{3} – 10^{3} + 3^{3}

Solution:

x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)

We know that a + b + c = 0 then a^{3} + b^{3} + c^{3} = 3ab

a + b + c = 7 + (-10) + 3

= 10 – 10

= 0

∴ 7^{3} – 10^{3} + 3^{3} = 3(7) (-10) (3)

= -630

(ii) 1 + \(\frac{1}{8}\) – \(\frac{27}{8}\)

Solution:

We know that a^{3} + b^{3} + c^{3} = 0 then a + b + c = 3abc

Question 14.

If 2x -3y – 4z = 0, then find 8x^{3} – 27y^{3} – 64z^{3}.

Solution:

We know x^{3} +y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)

x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} +y^{2} + z^{2} – xy – yz – zx) + 3xyz

8x^{3} – 27y^{3} – 64z^{3} = (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 (4x^{2} + 9y^{2} + 16z^{2} + 6xy – 12yz + 8xz) + 72xyz

= 72xyz

8x^{3} – 27y^{3} – 64z^{3} = 72xyz