How do you factor ## x^3 - 1##?

##x^3-1=(x-1)(x^2+x+1)=(x-1)(x-w)(x*w^2)## where ##w=-1/2+isqrt(3)/2##

Depending on whether you are factorising you polynomial in ##RR## or ##CC##.

Basically, you should use the factor theorem, so try reals that solves the equation, you should try all integer factors of -1, so +1 and -1.

By trying ##-1##, you see that ##(-1)^2-1=0##, so ##(x-1)## is a factor. Then you need to divide ##x^3-1## by ##x-1## using long division for example. ##(x^3-1)/(x-1)=(x^2+x+1)##.

And finally, you get that ##x^3-1=(x-1)(x^2+x+1)##.

Now you need to factorise a quadratic polynomial, so just use the formula, ##x={ -b+-sqrt(b^2-4ac)}/2##. And you get that ##w## and ##w^2## are the roots. So ##x^3-1=(x-1)(x-w)(x-w^2)##, and now you know that you're done because by D'Alembert's Theorem you know that polynomials in ##CC## have as many roots (not necessarily distinct) as the degree of the polynomial, in our case, 3.