Question 97039
Factor:
{{{x^3+8}}} Do you recognise this as the sum of two cubes?
{{{x^3+8 = (x)^3+(2)^3}}}...but of course!
The sum of two cubes can be factored thus:
{{{A^3+B^3 = (A+B)(A^2-AB+B^2)}}}
In your problem, A = x and B = 2, so...
{{{x^3+8 = (x+2)(x^2-2x+4)}}} and we're done!
Let's check the solution by multiplying the two factors, using FOIL.
{{{(x+2)(x^2-2x+4) = x^3-2x^2+4x+2x^2-4x+8}}} Simplify by combining like-terms.
{{{x^3+8 = x^3+8}}}