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.

Related questions
See all questions in Factor Polynomials Using Special Products
Impact of this question
1312 views around the world
You can reuse this answer
Creative Commons License