Question 109542
Factor completely:
{{{x^3-216}}} Should be able to recognise this as the difference of two cubes!
{{{(x)^3 - (6)^3}}} The difference of two cubes can be factored thus:
{{{A^3-B^3 = (A-B)(A^2+AB+B^2)}}} Applying this to your problem:
{{{(x)^3-(6)^3 = (x-6)(x^2+6x+36)}}} ...and this is as far as we can factor!