How do you factor #(x+y)^3 - 9(x^9 - y^9) #?

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)#

Explanation:

First, we will simplify #(x^9-y^9)#, since this is a difference of cubes. Differences of cubes fit the identity: #a^3-b^3=(a-b)(a^2+ab+b^2)#.

Thus:

#=(x+y)^3-9((x^3)^3-(y^3)^3)#

#=(x+y)^3-9(x^3-y^3)(x^6+x^3y^3+y^6)#

Now #(x^3-y^3)# 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)#

This is really the furthest this can be factored.

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