How do you factor out x cubed minus 1?

One obvious root of this is x = 1, so (x-1) is a factor.

Use long division of (x³ -1)/(x-1) and the quotient (x² + x + 1) will be another factor.
So (x³ -1) = (x-1)(x² + x + 1).

Remember that a cubic polynomial will always have 3 roots. It will either be 1 pure real root and 2 complex roots, or 3 pure real root. In this case (x² + x + 1) has two complex roots, and cannot be factored with real numbers.

In general, any odd-powered polynomial (3rd order, 5th order etc), will have at least one pure real root, and then pairs of either pure-real or complex roots.

Any even powered polynomial (2nd order, 4th order, etc) will have pairs of either pure-real, or complex roots.

Don't be fooled if you only find one root. Example: (x² + 2x + 1) actually has a double root, as it is factored into (x+1)(x+1).