How do you factor (x+y)^3 - 9(x^9 - y^9) (x+y)3−9(x9−y9)?

1 Answer
mason m
Aug 9, 2016

(x+y)^3-9(x-y)(x^2+xy+y^2)(x^6+x^3y^3+y^6)(x+y)3−9(x−y)(x2+xy+y2)(x6+x3y3+y6)

Explanation:

First, we will simplify (x^9-y^9)(x9−y9), since this is a difference of cubes. Differences of cubes fit the identity: a^3-b^3=(a-b)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2).

Thus:

=(x+y)^3-9((x^3)^3-(y^3)^3)=(x+y)3−9((x3)3−(y3)3)

=(x+y)^3-9(x^3-y^3)(x^6+x^3y^3+y^6)=(x+y)3−9(x3−y3)(x6+x3y3+y6)

Now (x^3-y^3)(x3−y3) can be factored in the same way:

=(x+y)^3-9(x-y)(x^2+xy+y^2)(x^6+x^3y^3+y^6)=(x+y)3−9(x−y)(x2+xy+y2)(x6+x3y3+y6)

This is really the furthest this can be factored.

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