How do you factor #a^3x^2 - 16a^3x + 64a^3 - b^3x^2 + 16b^3x - 64b^3#?

1 Answer
Apr 26, 2016

Answer:

#a^3x^2-16a^3x+64a^3-b^3x^2+16b^3x-64b^3#

#=(a-b)(a^2+ab+b^2)(x-8)^2#

Explanation:

Factor by grouping:

#a^3x^2-16a^3x+64a^3-b^3x^2+16b^3x-64b^3#

#=(a^3x^2-16a^3x+64a^3)-(b^3x^2-16b^3x+64b^3)#

#=a^3(x^2-16x+64)-b^3(x^2-16x+64)#

#=(a^3-b^3)(x^2-16x+64)#

The first factor can be factorised using the difference of cubes identity:

#(a^3-b^3) = (a-b)(a^2+ab+b^2)#

The second factor is a perfect square trinomial:

#x^2-16x+64 = (x-8)^2#

So:

#a^3x^2-16a^3x+64a^3-b^3x^2+16b^3x-64b^3#

#=(a-b)(a^2+ab+b^2)(x-8)^2#