How do you factor x^3 – 1?

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##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 elementising you polynomial in ##RR## or ##CC##.

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

By perplexing ##-1##, you see that ##(-1)^2-1=0##, so ##(x-1)## is a element. Then you deficiency to part ##x^3-1## by ##x-1## using crave dispersion for copy. ##(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 deficiency to elementise a quadratic polynomial, so righteous 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 recognize that you're done owing by D'Alembert's Theorem you recognize that polynomials in ##CC## feel as abundant roots (not necessarily independent) as the limit of the polynomial, in our occurrence, 3.

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