Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Students can download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

(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

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

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

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

(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

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

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

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 1

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

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)
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 2

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

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

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

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

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)]
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 3
= 63 – 3(6)
= 216 – 18
= 198

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

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:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 4
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:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 5
= 33 + 3(3)
= 27 + 9
= 36

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

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
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 6

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

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

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