How do you factor completely 8x^3 - 18xy^3?

Apr 10, 2016

$8 {x}^{3} - 18 x {y}^{3} = 2 x \left(4 {x}^{2} - 9 {y}^{3}\right)$

Explanation:

Note that both of the terms are divisible by $2 x$, so separate that out as a factor.

$8 {x}^{3} - 18 x {y}^{3} = 2 x \left(4 {x}^{2} - 9 {y}^{3}\right)$

This is as far as we can go since the degrees of the remaining terms are distinct prime numbers.

If the ${y}^{3}$ had been ${y}^{2}$ we would be able to factor the remaining expression as a difference of squares:

$4 {x}^{2} - 9 {y}^{2} = {\left(2 x\right)}^{2} - {\left(3 y\right)}^{2} = \left(2 x - 3 y\right) \left(2 x + 3 y\right)$