How do you factor #a^3 - (a-4)^3#?

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Answer 1 · 2016-02-15T15:00:17

#4(3a^2-12a+16)#

This is a difference of cubes, which factors into:

#x^3-y^3=(x-y)(x^2+xy+y^2)#

Here, we have #x=a# and #y=(a-4)#, which gives us a factorization of

#a^3-(a-4)^3#

#=(a-(a-4))(a^2+a(a-4)+(a-4)^2)#

We can continue to simplify.

#=(a-a+4)(a^2+a^2-4a+a^2-8a+16)#

#=4(3a^2-12a+16)#

This cannot be factored further (without the help of complex numbers).