Question

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

Answer 1

`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`