Question

please answer this question with solution ​

Answer 1

Answer:

How do you factor (x-y) ^3-8(x+y) ^3?

We have a nifty little rule that

a^3−b^3=(a−b)(a2+ab+b2)

Also, we know that 8=2^3

and a^3b^3=(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]

The big mess takes some care to manage the algebra without mistakes. We will have x^2s, xys, and y^2s involved.

7x^{2} + 6xy + 3y^{2}

So

= (−x−3y)[7x^2+6xy+3y^2]