How do you factor #a^3x^2 - 16a^3x + 64a^3 - b^3x^2 + 16b^3x - 64b^3#?
1 Answer
Apr 26, 2016
#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#
