Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3
Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3
Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 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
Solution:
(i) (3m + 5)2
Comparing (3m + 5)2 with (a + b)2 we have a = 3m and b = 5
(a + b)2 = a2 + 2 ab + b2
(3m + 5)2 = (3m)2 + 2 (3m) (5) + 52
= 32m2 + 30m + 25 = 9m2 + 30m +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 – 10p + 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)
= 22n2 + 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
Solution:
(i) (3 + m)3
Comparing (3 + m)3 with (a + b)3 we have a = 3; b = m
(a + b)3 = a2 + 3a2b + 3 ab2 + b3
(3 + m)3 = 33 + 3(3)2 (m) + 3 (3) m2 + m3
= 27 + 27m + 9m2 + m3 = m3 + 9 m2 + 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) 3 = a3 + 3a2b + 3ab2 + b3
(3p + 4q)3 = (3p)3 + 3(3p)2 (4q) + 3(3p)(4q)2 + (4q)3
= 33p3 +3 (9p2) (4q) + 9p (16q2) + 43q3
= 27p3 + 108p2q + 144pq2 + 64q3
(iv) (52)3 = (50 + 2)3
Comparing (50 + 2)3 with (a + b)3 we have a = 50 and b = 2
(a + b)3 = a3 + 3 a2b + 3 ab2 + b3
(50 + 2)3 = 503 + 3 (50)22 + 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 = (100 + 4)3
Comparing (100 + 4)3 with (a + b)3 we have a = 100 and b = 4
(a + b)3 = a3 + 3 a2b + 3 ab2 + b3
(100 + 4)3 = (100)3 + 3 (100)2 (4) + 3 (100) (4)2 + (4)3
= 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
Solution:
(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
(ii) (2x – 4y)3
Comparing (2x – 4y)3 with (a – b)3 we have a = 2x and b = 4y
(a – b)3 = a3 – 3a2b + 3ab3 – 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 + 3 ab (c)2 – c3
= a3b3 – 3(a2b2) c + 3abc2 – c3
= a3b3 – 3a2b2 c + 3abc2 – c3
(iv) (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
= 973 x3 y3 = (100 – 3)3 x3y3
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 (i) (5y + 1)(5y + 2)(5y + 3)
(ii) (p – 2)(p + 1)(p – 4)
Solution:
(i) (5y + 1) (5y + 2) (5y + 3)
Comparing (5y + 1) (5y + 2) (5y + 3) with (x + a) (x + b) (x + c) we have x = 5y ; a = 1; b = 2 and c = 3.
(x + a) (x + b) (x + c) = x3 + (a + b + c) x2 (ab + bc + ca) x + abc
= (5y)3 + (1 + 2 + 3) (5y)2 + [(1) (2) + (2) (3) + (3) (1)] 5y + (1)(2) (3)
= 53y3 + 6(52y2) + (2 + 6 + 3)5y + 6
= 1253 + 150y2 + 55y + 6
(ii) (p – 2)(p + 1)(p – 4) = (p + (-2))0 +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) = x3 + (a + b + c) x2 + (ab + be + 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
= p3 – 5p2 + 2p + 8
Question 5.
Find the volume of the cube whose side is (x + 1) cm.
Solution:
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 = (a3 + 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).
Solution:
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 – 2x2 – 5x + 6 units3
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