Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4

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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:
(i) (2x + 3y + 4z)²
(a + b + c)² ≡ a² + b² + c² + 2ab + 2bc + 2ca
∴ (2x + 3y + 4z)² = (2x)² + (3y)² + (4z)² + 2 (2x) (3y) + 2 (3y) (4z) + 2 (4z) (2x)
= 4x² + 9y² + 16z² + 12xy + 24yz + 16zx

(ii) (-p + 2q + 3r)²
(a + b + c)² ≡ a² + b² + c² + 2ab + 2bc + 2ca
(-p + 2q + 3r)² = (-p)² + (2q)² + (3r)² + 2(-p) (2q) + 2(2q) (3r) + 2 (-p) (3r)
= p² + 4q² + 9r² – 4pq +12qr – 6rp

(iii) (2p + 3)(2p – 4)(2p – 5)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + be + ca) x + abc
(2p + 3)(2p – 4)(2p – 5) = (2p)³ + (3 – 4 – 5) (2p)² + [(3 × – 4)² + (-4 × -5) + (-5 × 3)] 2p + 3 × – 4 × – 5
= 8p³ + (-6) (4p²) + [-12 + 20 + (-15)] 2p + 60
= 8p³ – 24p² + (-7)2p + 60
(2p + 3)(2p – 4)(2p – 5) = 8p³ – 24p² – 14p + 60

(iv) (3a + 1)(3a – 2)(3a + 4)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + bc + ca) x + abc
(3a + 1)(3a – 2)(3a + 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 – 8a – 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)
(ii) (2x + 3)(2x – 5)(2x – 6)
Solution:
(i) (x + 5)(x + 6)(x + 7)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + bc + ca) x + abc
Co-efficient of x² = a + b + c = 5 + 6 + 7 = 18
Co-efficient of x² = ab + bc + ca = (5 × 6) + (6 × 7) + (7 × 5)
= 30 + 42 + 35 = 107
Constant term = abc = 5 × 6 × 7
Co-efficient of constant term = 210

(ii) (2x + 3)(2x – 5)(2x — 6)
∴ Co-efficient of x² = 4 (a + b + c) = 4 (3 + (-5) + (-6))
= 4 × (-8) = -32
Co-efficient of x = 2 (ab + bc + ca)
= 2 [3 × (-5) + (-5) (-6) + (-6) (3)]
= 2[-15 + 30- 18] = 2 × (-3) = -6
Constant term = abc = 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) \(\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}\)
Solution:
(x + a)(x + b)(x + c) = x³ + 14x² + 59x + 70 …………….. (1)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + bc + ca) x + abc …………. (2)
(i) Comparing (1) & (2)
We get, a + b + c = 14
(ii) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{b c+a c+a b}{a b c}=\frac{59}{70}\)
(iii) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
= 14² – 2 (59) = 196 – 118 = 78
(iv) \(\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}=\frac{a^{2}+b^{2}+c^{2}}{a b c}=\frac{78}{70}=\frac{39}{35}\)

Question 4.
Expand
(i) (3a – 4b)³
(ii) (x + \(\frac{1}{y}\))³
Solution:
(i) (3a – 4b)³ We know that
(x – y)³ = x³ – 3x²y + 3xy² – y³
(3a-4b)³ = (3a)³ – 3 (3a)² (4b) + 3 (3a) (4b)² – (4b)³
= 27a³ – 108a²b + 144 ab² – 64b³

(ii) (x + \(\frac{1}{y}\))³
(x + y)³ ≡ x³ + 3x²y + 3xy² + y³
\(\left(x+\frac{1}{y}\right)^{3}=x^{3}+\frac{3 x^{2}}{y}+\frac{3 x}{y^{2}}+\frac{1}{y^{3}}\)

Question 5.
Evaluate the following by using identities:
(i) 98³
(ii) 1001³
Solution:
(i) 98² = (100 – 2)³
(a – b)³ ≡ a³ – 3a²b + 3ab² – b³
98³ = (100 – 2)³ = 100³ – 3 × 100² × + 3 × 100 × 2² – 2³
= 1000000 – 3 × 10000 × 2 + 300 × 4 – 8
= 1000000 – 60000 + 1200 – 8 = 1001200 – 60008 = 941192

(ii) 1001³ = (1000 + 1)³
(a + b)³ ≡ a³ + 3a²b + 3ab² + b³
(1000 + 1)³ = 1000³ + 3(1000²) × 1 + 3 × 1000 × 1² + 1³
= 1000,000,000 + 3,000,000 + 3000 + 1 = 1,003,003,001

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 and(xy + yz + zx) = 26
x² + y² + z² = (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 + 4b = 10, ab = 2
(3a + 4b)³ = (3a)³ + 3 (3a)² (4b) + 3 (3a) (4b)² + (4b)³
(27a³ + 64b³) = (3a + 4b)³ – 3 (3a) (4b) (3a + 4b)
∵ x³ + y³ = (x + y)³ – 3xy – (x + y)
= 10³ – 36 ab (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 – y³ = (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:
a³ + b³ = (a + b)³ – 3ab (a + b)

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:

Question 11.
If (y – \(\frac{1}{y}\))³ = 27, then find the value of y³ – \(\frac{1}{y^{3}}\)
Solution:

Question 12.
Simplify : (i) (2a + 36 + 4c)(4a² + 9b² + 16c² – 6ab – 12bc – 12bc – 8ca)
(ii) (x – 2y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
Solution:
(i) (2a + 36 + 4c)(4a² + 9b² + 16c² – 6ab – 12bc – 12bc – 8ca)
We know that
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ × 3 abc
∴ (2a + 36 + 4c) (4a² + 9b² + 16 c² – 6 ab – 12 bc – 8 ca)
= (2a)³ + (3b)³ + (4c)³ – 3 × 2a × 36 × 4c
= 8a³ + 27b³ + 64c³ – 72 abc

(ii) (x – 2,y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ – 3abc .
∴ (x – 2y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
= x³ + (-2y)³ + (3z)³ – 3 × x × (-2y) (3z)
= x³ – 8y³ + 27z³ + 18 xyz

Question 13.
By using identity evaluate the following:
(i) 7³ – 10³ + 3³
(ii) 1 + \(\frac{1}{8}-\frac{27}{8}\)
Solution:
(i) 7³ – 10³ + 3³
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ – 3 abc
If a + b + c = 0, then a³ + b³ + c³ = 3 abc
∴ 7 – 10 + 3 = 0
⇒ 7³ – 10³ + 3³ = 3 × 7 × – 10 × 3
= 9 × -70 = -630

Question 14.
If 2x – 3y – 4z = 0, then find 8x³ – 27y³ – 64z³.
Solution:
If 2x – 3y – 4z = 0 then 8x³ – 27y³ – 64z³ = ?
If x + y + z = 0 then x³ + y³ + z³ = 3xyz
8x³ – 21 y³ – 64z³ = (2x)³ + (-3y)³ + (-4z)³
= 3 × 2x × -3y × – 4z = 72 xyz