If α and β are the roots of $x^2-3x-4=0$, the equation whose roots are α³ and β³ is:

With α + β = 3 and αβ = −4, the cubes add to α³ + β³ = $(\alpha +\beta {)}^3-3$αβ(α + β) = 27 + 36 = 63 and multiply to $(\alpha \beta {)}^3=-64$. The required equation is $x^2-63x-64=0$. (Using only $(\alpha +\beta {)}^3=27$ gives the first option; the sign of the product −64 is dropped in the third.) The original roots are 4 and −1, so the new roots are 64 and −1, which indeed add to 63 and multiply to −64.