Question

factorise:
(x-y)^3+8(x+y)^3

Answer 1

Answer:

a3−b3=(a−b)(a2+ab+b2)  

Also, we know that 8=23

and a3b3=(ab)3

So (x−y)3−8(x+y)3

= (x−y)3−(2x+2y)3

=[(x−y)−(2x+2y)][(x−y)2+(x−y)(2x+2y)+(2x+2y)2]

= (−x−3y)[bigmess]

Step-by-step explanation:

Answer 2

ANSWER:

To Factorize:

• (x - y)^3 + 8(x + y)^3

Solution:

We are given that,

$\implies (x-y)^3+8(x+y)^3$

We can re-write it as,

$\implies (x-y)^3+2^3(x+y)^3$

$\implies (x-y)^3+[2(x+y)]^3$

So,

$\implies (x-y)^3+(2x+2y)^3$

We know that,

$\hookrightarrow a^3+b^3=(a+b)(a^2+b^2-ab)$

In this question, a = (x - y) and b = (2x + 2y)

So,

$\implies (x-y)^3+(2x+2y)^3$

$\implies (x-y+2x+2y)[(x-y)^2+(2x+2y)^2-(x-y)(2x+2y)]$

$\implies (3x+y)[x^2+y^2-2xy+4x^2+4y^2+8xy-2x^2-2xy+2xy+2y^2]$

On rearranging and grouping,

$\implies (3x+y)[(x^2+4x^2-2x^2)+(y^2+4y^2+2y^2)+(-2xy+8xy-2xy+2xy)]$

Hence,

$\implies \bf(3x+y)[3x^2+7y^2+6xy]$