Question

help me please.... factorisation​

Answer 1

Step-by-step explanation:

Solution:-

13)

Given expression is (x+1)^6-(x-1)^6

It can be written as

=> [(x+1)^2]^3 - [(x-1)^2]^3

This is in the form of a^3-b^3

Where a = (x+1)^2 and b = (x-1)^2

We know that

a^3 - b^3 = (a-b)(a^2+ab+b^2)

=>[ (x+1)^2-(x-1)^2][(x+1)^4+(x+1)^2(x-1)^2+(x-1)^4]

=> (x+1+x-1)(x+1-x+1)[(x+1)^4+(x+1)^2(x-1)^2+(x-1)^4]

=> (2x)(2)[(x+1)^4+(x+1)^2(x-1)^2+(x-1)^4]

=> (4x)[(x+1)^4+(x+1)^2(x-1)^2+(x-1)^4]

(x+1)^6-(x-1)^6

=(4x)[(x+1)^4+(x+1)^2(x-1)^2+(x-1)^4]

14)

Given expression is 3a^7b - 81a^4b^4

=> 3a^4b(a^3-27b^3)

=> 3a^4b(a^3-(3b)^3)

=> 3a^4b(a-3b)(a^2+3ab+9b^2)

Since a^3 - b^3 = (a-b)(a^2+ab+b^2)

3a^7b - 81a^4b^4 =

3a^4b(a-3b)(a^2+3ab+9b^2)

15)

Given expression is

9(x+y)^2-24(x^2-y^2)+16(x-y)^2

=> 9(x^2+2xy+y^2)-24x^2+24y^2+16(x^2-2xy+y^2)

=> 9x^2+18xy+9y^2-24x^2+24y^2+16x^2-32xy+16y^2

=> (9x^2+16x^2-24x^2)+(18xy-32xy) + ( 9y^2+24y^2+16y^2)

=> (25x^2-24y^2)+(18xy-32xy) +(33y^2+16y^2)

=> x^2 +(-14xy)+ 49y^2

=> x^2-14xy+49y^2

=> (x)^2-2(x)(7y)+(7y)^2

Since (a-b)^2 = a^2-2ab+b^2

=>(x-7y)^2

=> (x-7y)(x-7y)

9(x+y)^2-24(x^2-y^2)+16(x-y)^2

= (x-7y)(x-7y)

Used formulae:-

• a^3 - b^3 = (a-b)(a^2+ab+b^2)

• (a-b)^2 = a^2-2ab+b^2

• (a+b)^2=a^2+2ab+b^2

• (a+b)(a-b)=a^2-b^2