Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.3 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.3

Question 1.
Expand
(i) (3m + 5)2
(ii) (5p – 1)2
(iii) (2n – 1)(2n + 3)
(iv) 4p2 – 25q2
(i) (3m + 5)2
Comparing (3m + 5)2 with (a + b)2 we have a = 3m and b = 5
(a + b)2 = a2 + 2ab + b2
(3m + 5)2 = (3m)2 + 2(3m) (5) + 52
= 32 m2 + 30 m + 25
= 9m2 + 30 m + 25 (ii) (5p – 1)2
Comparing (5p – 1)2 with (a – b)2 we have a = 5p and b = 1
(a – b)2 = a2 – 2ab + b2
(5p – 1)2 = (5p)2 – 2(5p)(1) + 12
= 52p2 – 10 p + 1
= 25p2 – 10p + 1

(iii) (2n – 1)(2n + 3)
Comparing (2n – 1) (2n + 3) with (x + a)(x + b) we have a = -1; b = 3
(x + a) (x + b) = x2 + (a + b)x + ab
(2n + (-1)) (2n + 3) = (2n)2 + (-1 + 3)2n + (-1) (3)
= 22 n2 + 2(2n) – 3 = 4n2 + 4n – 3

(iv) 4p2 – 25q2 = (2p)2 – (5q)2
Comparing (2p)2 – (5q)2 with a2 – b2 we have a = 2p and b = 5q
(a2 – b2) = (a + b)(a – b)
= (2p + 5q)(2p – 5q) Question 2.
Expand
(i) (3 + m)3
(ii) (2a + 5)3
(iii) (3p + 4q)3
(iv) (52)3
(v) (104)3
(i) (3 + m)3
Cornparing (3 + m)3 with (a + b)3 we have a = 3 ; b = m
(a + b)3 = a2 + 3a2b + 3ab2 + b3
(3 + m)3 = 33 + 3(3)2 (m) + 3(3)m2 + m3
= 27 + 27m + 9m2 + m3
= m3 + 9m2 + 27m + 27

(ii) (2a + 5)3 =
Comparing (2a + 5)3 with (a + b)3 we have a = 2a, b = 5
(a + b)3 = a3 + 3a2b + 3ab2 + b3
= (2a)3 + 3(2a)2 5 + 3 (2a) 52 + 53
= 23a3 + 3(22a2)5 + 6a(25) + 125
= 8a3 + 60a2 + 150a + 125 (iii) (3p + 4q)3
Comparing (3p + 4q)3 with (a + b)3 we have a = 3p and b = 4q
(a + b) = a3 + 3a2b + 3ab2 + b2
(3p + 4q)3 = (3p)3 + 3(3p)2 (4q) + 3(3p)(4q)2 + (4q)3
= 33p3 + 3(9p2)(4q) + 9p(16q2) – 43q3
= 27p3 + 108p2q + 144pq3 + 64q3

(iv) (52)3
(52)3 = (50 + 2)3
Comparing (50 + 2)3 with (a + b)3 we have a = 50 and b =2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(50 + 2)3 = 503 + 3 (50)2 2 + 3 (50)(2)2 + 23
523 = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
523 = 1,40,608

(v) (104)3
(104)3 = (100 + 4)3
Comparing (100 + 4)3 with (a + b)3 we have a = 100 and b = 4
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(100 + 4)3 = (100)3 + 3 (100)2 (4) + 3 (100) (4)2 + 43
= 10,00,000 + 3(10000)4 + 300(16) + 64
= 10,00,000 + 1,20,000 + 4,800 + 64= 11,24,864 Question 3.
Expand
(i) (5- x)3
(ii) (2x – 4y)3
(iii) (ab – c)3
(iv) (48)3
(v) (97xy)3
(i) (5- x)3
Comparing (5 – x)3 with (a – b)3 we have a = 5 and b = x
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(5 – x)3 = 53 – 3(5)2(x) + 3(5)(x2) – x3
= 125 – 3(25)(x) + 15x2 – x3
= 125 – 75x + 15 x2 – x3 (ii) (2x – 4y)3
Comparing (2x – 4y)3 with (a – b)3 we have a = 2x and b = 4y
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(2x – 4y)3 = (2x)3 – 3(2x)2 (4y) + 3(2x) (4y)2 – (4y)3
= 23x3 – 3(22x2)(4y) + 3(2x) (42y2) – (43y3)
= 8x3 – 48x2y + 96xy2 – 64y3

(iii) (ab – c)3
Comparing (ab – c)3 with (a – b)3 we have a = ab and b = c
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(ab – c)3 = (ab)3 – 3(ab)2 c + 3ab (c)2 – c3
= a3b3 – 3(a2b2) c + 3abc2 – c3
= a3b3 – 3a2b2c + 3abc2 – c3

(iv) (48)3
(48)3 = (50 – 2)3
Comparing (50- 2)3 with (a – b)3 we have a = 50 and b = 2
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(50 – 2)3 = (50)3 – 3(50)2 (2) + 3 (50)(2)2 – 23
= 1,25,000 – 15000 + 600 – 8
= 1,10,000 + 592
= 1,10,592 (v) (97xy)3
(97xy)3 = 973 x3 y3 = (100 – 3) x3y3 … (1)
Comparing(100 – 3)3 with (a – b)3 we have a = 100, b = 3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(100 – 3)3 = (100)3 – 3(100)2 (3) + 3 (100)(3)2 – 33
973 = 10,00,000 – 90000 + 2700 – 27
973 = 910000 + 2673
973 = 912673
97x3y3 = 912673x3y3

Question 4.
Simplify (p – 2)(p + 1)(p – 4)
(p – 2)(p + 1)(p – 4) = (p + (-2)) (p + 1) (p + (-4))
Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2;
b = 1 ; c = -4.
(x + a)(x + b)(x + c) = x2 + (a + b + c) x2 + (ab + bc+ ca)x + abc
= p3 + (-2 + 1 + (-4)) p2 + (-2)( 1) + (1)(-4) + (-4) (-2))p + (-2) (1) (-4)
= p3 +(-5)p2 + (-2 + (-4) + 8)p + 8
= p2 – 5p2 + 2p + 8 Question 5.
Find the volume of the cube whose side is (x + 1) cm
Given side of the cube = (x + 1) cm
Volume of the cube = (side)3 cubic units = (x + 1)3 cm3
We have (a + b)3 = (a33 + 3a2b + 3ab2 + b3) cm3
(x + 1)3 = (x3 + 3x2(1) + 3x(1)2 + 13)cm3
Volume = (x3 + 3x2 + 3x + 1) cm3

Question 6.
Find the volume of the cuboid whose dimensions are (x + 2),(x – 1) and (x – 3)
Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)
∴ Volume of the cuboid = (l × b × h) units3
= (x + 2) (x – 1) (x – 3) units3
We have (x + a)(x + b) (x+c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
∴ (x+2) (x- 1) (x-3) = x3 + (2 – 1 – 3)x2 + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2) (-1) (-3)
= x3 – 2x2 + (-2 + 3 – 6)x + 6
Volume = x3 – 2x3 – 5x + 6 units3 Objective Type Questions

Question 7.
If x2 – y2 = 16 and (x + y) = 8 then (x – y) is ________
(A) 8
(B) 3
(C) 2
(D) 1
(C) 2
Hint:
x2 – y2 = 16
(x + y) (x – y) = 16
8 (x – y) = 16
(x – y) = $$\frac { 16 }{ 8 }$$ = 2

Question 8.
$$\frac{(a+b)\left(a^{3}-b^{3}\right)}{\left(a^{2}-b^{2}\right)}$$ = _________
(A) a2 – ab + b2
(B) a2 + ab + b2
(C) a2 + 2ab + b2
(D) a22 – 2ab + b2
(B) a2 + ab + b2
Hint: = a2 + ab + b2 Question 9.
(p + q)(p2 – pq + q2) is equal to __________
(A) p3 + q3
(B) (p + q)3
(C) p3 – q3
(D) (p – q)3
(A) p3 + q3
Hint:
a3 + b3 = (a + b) (a2 – ab + b2)

Question 10.
(a – b) = 3 and ab = 5 then a3 – b3 = __________
(A) 15
(B) 18
(C) 62
(D) 72
(D) 72
Hint:
(a – b) = 3
(a – b)2 = 32
a2 + b2 – 2ab = 9
a2 + b2 – 2(5) = 9
a2 + b2 = 9 + 10
a2 + b2 = 19
a3 – b3 = (a – b)(a2 + ab + b2) = 3(19 + 5)
= 3(24) = 72 Question 11.
a3 + b3 = (a + b)3 _________
(A) 3a(a + b)
(B) 3ab(a – b)
(C) -3ab(a + b)
(D) 3ab(a + b)