# what is the complete factorization of p^3-8?

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The difference of two cubes is `x^3-a^3=(x-a)(x^2+xa+a^2)`

So `p^3-8=(p-2)(p^2+2p+4)`

Factoring `p^2+2p+4` using the quadradic formula is

`p=(-2+-sqrt(4-16))/2=-1+-sqrt(-12)/2=-1+-2isqrt(3)/2=-1+-isqrt(3)`

So the complete factorization is

`p^3-8=(p-2)(p+1+isqrt(3))(p+1-isqrt(3))`