Algebra Chapter 6 Fraction By Formulae
Fraction By Formulae Introduction
A very important topic in algebra is to resolve an expression into factors. In the previous chapter, we studied some of the very important formulae. In this chapter, our aim is to apply those formulae to resolve algebraic expressions into factors.
The sum and difference of two cubes
In the previous chapter we have seen that, (a + b) (a2 – ab + b²) = a3 + b 3 and
(a – b) (a2 + ab + b2) = a3 – b3.
Thus, we may conclude that the two factors of a3 + b3 are (a + b) and (a2 – ab + b2),
and also the two factors of a3 – b3 are (a – b) and (a2 + ab + b2).
We may also verify the above two formulae from the reverse direction as shown below
a3 + b3 = a3 + a2b – a2b – ab2 + ab2 + b3
= a2(a + b) – ab(a + b) + b2(a + b)
= (a + b) (a2 – ab + b2).
a3 – b3 = a3 – a2b + a2b – ab2 + ab2 – b3
= a2(a – b) + ab(a – b) + b2(a – b)
= (a – b) (a2 + ab + b2)
Read And Learn More WBBSE Solutions For Class 8 Maths
Algebra Chapter 6 Fraction By Formulae Some Examples of Factors
Example 1
Factorize: x3 + 64.
Solution :
Given:
x3 + 64
x3 + 64 = (x)3 + (4)3
= (x + 4) {(x)2 – x.4 + (4)2}
= (x + 4) (x2 – 4x + 16)
x3 + 64 = (x + 4) (x2 – 4x + 16)
Example 2
Factorize: 8a3 – 27b3.
Solution :
Given:
8a3 – 27b3.
8a3 – 27b3 = (2a)3 – (3b)3
= (2a – 3b) {(2a)2 + 2a.3b + (36)2}
= (2a – 3b) (4a2 + 6ab + 9b2)
8a3 – 27b3 = (2a – 3b) (4a2 + 6ab + 9b2)
WBBSE Class 8 Fraction by Formulae Notes
Example 3
Factorize: a6 – b6.
Solution :
Given:
a6 – b6
a6 – b6 = (a3)2 – (b3)2
= (a3 + b3) (a3 – b3)
= (a + b) (a2-ab+b2) (a-b) (a2+ab + b2)
a6 – b6 = (a + b) (a2-ab+b2) (a-b) (a2+ab + b2)
Example 4
Factorize: 3x3 + 375.
Solution :
Given:-
3x3 + 375
3x3 + 375 = 3(x3 + 125) = 3{(x)3 + (5)3}
= 3(x + 5) {(x)2 – x.5 + (5)2}
= 3(x + 5) (x2 – 5x + 25)
3x3 + 375 = 3(x + 5) (x2 – 5x + 25)
Example 5
Factorize: a4b- ab4.
Solution :
Given: a4b- ab4
a4 b- ab4
= ab(a3 – b3)
= ab(a – b) (a2 + ab + b2)
a4b- ab4 = ab(a – b) (a2 + ab + b2)
Understanding Fraction Operations in Algebra
Example 6
Factorize : a3 + 3a2 b+ 3ab2 + 2b3.
Solution :
Given:
a3 + 3a2 b + 3ab2 + 2b3
a3 + 3a2 b + 3ab2 + 2b3
= a3 + 3a2 b+ 3ab2 + b3 + b3
= (a + b)3 + (b)3
= (a + b + b) {(a + b)2 – (a + b).b + (b)2}
= (a + 2b) (a² + 2ab + b2 – ab – b2 + b2)
= (a + 2b) (a2 + ab + b2)
a3 + 3a2 b + 3ab2 + 2b3 = (a + 2b) (a2 + ab + b2)
Example 7
Factorize : 8a3 + 36a2b + 54ab2 + 27b3.
Solution :
Given:
8a3 + 36a2 b + 54ab2 + 27b3
8a3 + 36a2 b + 54ab2 + 27b3
= (2a)3 + 3 (2a)2 x 3b + 3 x 2a (3b)2 + (3b)3
= (2a + 3b)3
= (2a + 3b) (2a + 3b) (2a + 3b)
8a3 + 36a2 b + 54ab2 + 27b3 = (2a + 3b) (2a + 3b) (2a + 3b)
Example 8
Factorize: 35 – a3 + 6a2 – 12a.
Solution :
Given:
35 – a3 + 6a2 – 12a
35 – a3 + 6a2 – 12a
= 27 + 8 – a3 + 6a2 – 12a
= 27 – (a3 – 6a2 + 12a – 8)
= 27 – {(a)3 – 3.(a)2.2 + 3.a.(2)2 – (2)3}
= (3)3 – (a – 2)3
= {3 – (a – 2)} {(3)2 + 3.(a – 2) + (a – 2)2}
= (3 – a + 2) (9 + 3a – 6 + a² – 4a + 4)
= (5 – a) (a2 – a + 7)
35 – a3 + 6a2 – 12a = (5 – a) (a2 – a + 7)
Step-by-Step Guide to Adding and Subtracting Fractions
Example 9
Factorize : 8(a + b)3 + 27(6 + c)3.
Solution :
Given:
8(a + b)3 + 27(6 + c)3.
8(a + b)3 + 27(b + c)3 = {2(a + b)}3 + {3(b + c)}3
= (2a + 2b)3 + (3b + 3c)3
= (2a + 2b + 3b + 3c) {(2a + 2b)2 – (2a + 2b) (3b + 3c) + (3b + 3c)2}
= (2a + 5b + 3c) {(2a)2 + 2 x 2a x 2b + (2b)2 – (6ab + 6ac + 6b2 + 6bc) + (3b)2 + 2 x 3b x 3c + (3c)2}
= (2a + 5b + 3c) {4a2 + 8ab + 462 – 6ab – 6ac – 6b2 – 6bc + 9b2 + 18bc + 9c2}
= (2a + 56 + 3c) (4a² + 7b² + 9c2 + 2ab + 12bc – 6ac)
8(a + b)3 + 27(6 + c)3.= (2a + 56 + 3c) (4a² + 7b² + 9c2 + 2ab + 12bc – 6ac)
Example 10
Factorize : x3 + y3 – x(x2 – y2) + y(x + y)2.
Solution :
Given
x3 + y3 – x(x2 – y2) + y(x + y)2.
x3 + y3 – x(x2 – y2) + y(x + y)2
= (x +y)(x2-xy +y2) -x(x +y)(x-y) +y(x+y)
= (x+y) {(x²-xy+y³)-x(x-y)+y(x+y)}
= (x + y) {x2– xy + y2 – x2 + xy + xy + y2}
= (x +y) (xy + 2y2)
= (x+y) y(x + 2y)
= y(x + y) (x + 2y)
x3 + y3 – x(x2 – y2) + y(x + y)2. = y(x + y) (x + 2y)
Practice Problems on Fractions for Class 8
Example 11
Factorize : x3 + 9x2 + 21x + 26.
Solution :
Given:
x3 + 9x2 + 21x + 26.
x3 + 9x2 + 21 x + 26
= (x)3 + 3(x)2 + 3x (3)2 + (3)3 – 1
= (x + 3)3 – (1)3
= (x + 3 – 1) {(x + 3)2 + (x + 3)1 + (1)2}
= (x + 2) {x2 + 3x + 9 + x + 3 + 1} .
= (x + 2) (x2 + 7x + 13)
x3 + 9x2 + 21x + 26. = (x + 2) (x2 + 7x + 13)
Example 12
Factorize : x3 – 6xy + 12x2 y- 8y3 – z + 3z2 – 3z + 1.
Solution:
Given:
x3 – 6x2y + 12xy2 – 8y3 – z3 + 3z2 – 3z + 1
x3 – 6x2y + 12xy2 – 8y3 – z3 + 3z2 – 3z + 1
= (x)3 – 3(x)2 2y + 3x(2y)2 – (2y)3 – {(z)3 – 3(z)2 x 1 + 3z(1)2 – (1)3}
= (x – 2y)3 – (z-1)3
= {(x-2y) – (z-1)}{(x-2y)2 + (x-2y) . (z-1) + (z-1)2}
= (x – 2y – z +1)(x2 + 4y2 + z2 – 4xy + zx – 2yz – x + 2y – 2z +1)
x3 – 6x2y + 12xy2 – 8y3 – z3 + 3z2 – 3z + 1 = (x – 2y – z +1)(x2 + 4y2 + z2 – 4xy + zx – 2yz – x + 2y – 2z +1)
Examples of Multiplying and Dividing Fractions
Example 13
Factorize : 16a3 – 54(a – b)3.
Solution :
Given
16a3 – 54(a – b)3
16a3 – 54(a – b)3
= 2[8a3 – 27(a – b)3]
= 2[(2a)3 – {3(a – b)}3]
= 2{(2a)3 – (3a – 3b)3}
= 2(2a – 3a + 3b){(2a)2 + 2a(3a – 3b) + (3a – 3b)2}
= 2(36 – a)(4a2 + 6a2 — 6ab + 9a2 – 18ab + 9b2)
= 2(36 – a)(19a2 — 24ab + 9b2)
16a3 – 54(a – b)3 = 2(36 – a)(19a2 — 24ab + 9b2)
Example 14
Factorize: 8 – a3 + 3a2b – 3ab2 + b3.
Solution :
Given:
8 – a3 + 3a2b – 3ab2 + b3
8 – a3 + 3a2b- 3ab2 + b3 = 8 – (a3 – 3a2 b+ 3ab2 – b3)
= (2)3 – (a – b)3
= (2 – a + b){(2)2 + 2(a – b) + (a – b)2}
= (2 – a + b)(4 + 2a — 2b + a2 – 2ab + b2)
= (2 – a + b)(a2 — 2ab + b2 + 2a – 2b + 4)
8 – a3 + 3a2b – 3ab2 + b3 = (2 – a + b)(a2 — 2ab + b2 + 2a – 2b + 4)
Conceptual Questions on Fractions and Their Applications
Example 15
Factorize : m3 – n3 – m(m2 – n2) + n(m – n.)2.
Solution :
Given
m3 – n3 – m(m2 – n2) + n(m – n.)2
m3 – n3 – m(m2 – n2) + n(m – rc)2
= (m – n)(m2 + mn + n2) – m(m + n)(m – n) + n(m – n)2
= (m – n)(m2+ mn + n2– m2– mn+mn – n2)
= (m – n)(mn)
= mn(m – n)
m3 – n3 – m(m2 – n2) + n(m – n.)2 = mn(m – n)
Example 16
Factorize : 8x3+12x2+6x – y3+9y2-27y + 28.
Solution :
Given
8x3+12x2+6x – y3+9y2-27y + 28.
8x3+ 12X2 + 6x – y3 + 9y2 – 27y + 28
= 8x3 + I2x2 + 6x + 1 -y3 + 9y2 – 27y + 27
= (2x)3 + 3 (2x)2 x 1 + 3 x 2x (1)2 + 1- {(y)3-3y2x3 + 3y(3)2-(3)3}
= (2x + 1)3 – (y – 3)3
= {(2x + 1) – (y – 3)}{(2x + 1)2 + (2x + 1)(y – 3) + (y – 3)2}
= (2x + 1 – y + 3)(4x2 + 4x + 1 + 2xy – 6x + y – 3 + y2 – 6y + 9)
= (2x – y + 4 x 4x2 + 2xy + y2 -2x- 5y + 7)
8x3+12x2+6x – y3+9y2-27y + 28. = (2x – y + 4 x 4x2 + 2xy + y2 -2x- 5y + 7)
Example 17
Factorize : x3-9y3– 3xy(x-y)
Solution :
Given
x3-9y3– 3xy(x-y)
x3 – 9y3 – 3xy (x – y)
= [x3-y3– 3xy (x – y)] – 8y3
= (x- y)3 – 8y3
= (x- y)3 – (2y)3
= {(x – y)-2y} . {(x – y)2 + (x – y) . 2y + (2y)2}
= (x- 3y).(x2 – 2xy + y2 + 2xy – 2y2 + 4y2)
= (x-3y). (x2 + 3y2)
x3-9y3– 3xy(x-y) = (x-3y). (x2 + 3y2)
Example 18
Factorize : a3 – 9b3 + (a + b)3
Solution :
Given
a3 – 9b3 + (a + b)3
a3 – 9b3+ (a + b)3
= a3 – b3 + (a + b)3 – 8b3
= a3 – b3 +(a + b)3 – (2b)3
= (a – b) (a2 + ab + b2) + {(a + b) – 2b}. {(a + b)2 + (a + 6).2b + (2b)2}
= (a – b) (a2 + ab + b2) + (a – b) (a2 + 4ab + 7b2) = (a – b)(a2 + ab + b2 +a2 + 4ab + 7b2)
= (a – b).(2a2 + 5ab + 3b2)
a3 – 9b3 + (a + b)3 = (a – b).(2a2 + 5ab + 3b2)
Example 19
Resolve into factors : 2x3 – 3x2 + 3x – 1
Solution :
Given:
2x3 – 3x2 + 3x – 1
2x3 – 3x2 + 3x – 1 = x3 + x3 – 3x2 + 3x – 1
= x3 + (x – l)3
= {x + (x – 1)} {x2 – x.(x – 1)+ (x – l)2}
= (2x – 1) (x2 – x2 + x + x2 – 2x + 1)
= (2x – 1) (x2 – x + 1)
2x3 – 3x2 + 3x – 1 = (2x – 1) (x2 – x + 1)
Example 20
Resolve into factors : a3 – 12a – 16
Solution :
Given:
a3 – 12a – 16
a3 – 12a – 16
= a3 + 8 – 12a – 24
= a3 + 23 – 12(a + 2)
= (a + 2) (a2 – 2a +22) – 12(a + 2)
= (a + 2) (a2 – 2a + 4 – 12)
= (a + 2) (a2 – 2a – 8)
= (a + 2). (a2 – 4a + 2a – 8)
= (a + 2) {a(a – 4) + 2 (a – 4)}
= (a + 2) (a – 4) ( a + 2)
a3 – 12a – 16 = (a + 2) (a – 4) ( a + 2).