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

Apr 26, 2016

${a}^{3} {x}^{2} - 16 {a}^{3} x + 64 {a}^{3} - {b}^{3} {x}^{2} + 16 {b}^{3} x - 64 {b}^{3}$

$= \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right) {\left(x - 8\right)}^{2}$

#### Explanation:

Factor by grouping:

${a}^{3} {x}^{2} - 16 {a}^{3} x + 64 {a}^{3} - {b}^{3} {x}^{2} + 16 {b}^{3} x - 64 {b}^{3}$

$= \left({a}^{3} {x}^{2} - 16 {a}^{3} x + 64 {a}^{3}\right) - \left({b}^{3} {x}^{2} - 16 {b}^{3} x + 64 {b}^{3}\right)$

$= {a}^{3} \left({x}^{2} - 16 x + 64\right) - {b}^{3} \left({x}^{2} - 16 x + 64\right)$

$= \left({a}^{3} - {b}^{3}\right) \left({x}^{2} - 16 x + 64\right)$

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

$\left({a}^{3} - {b}^{3}\right) = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

The second factor is a perfect square trinomial:

${x}^{2} - 16 x + 64 = {\left(x - 8\right)}^{2}$

So:

${a}^{3} {x}^{2} - 16 {a}^{3} x + 64 {a}^{3} - {b}^{3} {x}^{2} + 16 {b}^{3} x - 64 {b}^{3}$

$= \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right) {\left(x - 8\right)}^{2}$