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

1 Answer
Apr 10, 2016

Answer:

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

Explanation:

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?