Generalization of $x^3-2=(x-\sqrt[3]{2})(x-\sqrt[3]{2}w)(x-\sqrt[3]{2}w^2)$

Question

Note that \begin{align} x^3-1 = (x-1)(x^2+x+1) =0 \end{align} has solution of $x=1, w, w^2$ where $w^2+w+1=0$,

and further i know

$x^3-2=(x-\sqrt[3]{2})(x-\sqrt[3]{2}w)(x-\sqrt[3]{2}w^2)$

can it be generalized to any number? For example

$x^3-q=(x-\sqrt[3]{q})(x-\sqrt[3]{q}w)(x-\sqrt[3]{q}w^2)$

like above? How one can prove this?

For $x^3-2$ by direct expansion of R.H.S I notice left and right sides are indeed same. And i think same thing holds for arbitrary number $q$.

But is there any other way to factorized this?

Answer 1

For solving $x^3-q=0$ substitute $x=q^{\frac{1}{3}}y$

then the equation becomes $y^3=1$ whose solutions are $1,\omega,\omega^2$

Hence $x=q^{\frac{1}{3}},q^{\frac{1}{3}}\omega,q^{\frac{1}{3}}\omega^2$

Answer 2

The formula that you state is sound. Just expand the bracket & use $\color{blue}{(1+w+w^2)=0}$ & $w^3=1$. \begin{eqnarray*} (x-\sqrt[3]{q})(x-\sqrt[3]{q}w)(x-\sqrt[3]{q}w^2) &=& x^3 -x^2\sqrt[3]{q}(\color{blue}{1+w+w^2})+ x(\sqrt[3]{q})^2(\color{blue}{1+w+w^2}) +q \\&=& x^3-q. \end{eqnarray*}