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

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Answer 1 · 2016-04-10T07:46:31

$8x^3-18xy^3 = 2x(4x^2-9y^3)$

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

$8x^3-18xy^3 = 2x(4x^2-9y^3)$

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:

$4x^2-9y^2 = (2x)^2-(3y)^2 = (2x-3y)(2x+3y)$

Was this a typo in the question?

How do you factor completely 8x^3 - 18xy^38x^3 - 18xy^3?