find the factorisation of (x–y)^3+8(x+y)^3
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find the factorisation of (x–y)^3+8(x+y)^3
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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]
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)[7x2+6xy+3y2]
Step-by-step explanation:
(x-y) ^3-8(x+y) ^3?
We have a nifty little rule that
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]