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).
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(x - 1)(x - 1)(x - 1) - (x + 1)(x + 1)(x + 1)
x(x + 1)(x -1)
(x - 5)(x^2 + 1)
x³ - x² + x - 3 (x² + 1)(x - 1) - 2
x(x - 19)(x - 1)