<s>[INST] Lagrange Method of Solving Cubic Equations Let $K$ be a field (for simplicity, one may assume $K\subseteq\mathbb{C}$), and let $d\in K$. Denote $w=\sqrt[3]{d}$, and $\zeta=(-1+\sqrt{-3})/2$. If $f(x)=x^3+ax^2+bx+c$ is the minimal polynomial of $x_{1}=\alpha+\beta w+\gamma w^2$, show that:


*

*The other two roots of $f$ are $x_{2}=\alpha+\zeta\beta w+\zeta^2\gamma w^2$ and $x_{3}=\alpha+\zeta^2\beta w+\zeta\gamma w^2$.

*$\alpha = (x_1+x_2+x_3)/3$, $\beta w=(x_1+\zeta x_2 +\zeta^2 x_3)/3$ and $\gamma w^2=(x_1+\zeta^2x_2+\zeta x_3).$
I tried expanding the term $f(x_1)$ (see below), using the fact that $1+\zeta+\zeta^2=0$ and $\zeta^3=1$, however I found that under the transformation $\beta \mapsto\zeta\beta, \gamma \mapsto\zeta^2\gamma$ the coefficients $a_j$ of $w^j$ ($0\le j\le 6$) are multiplied by $\zeta^j$ too. How can I show that this transformation does not change the result and deduce that $f(x_2)=f(x_3)=0$, without explicitly solving the original equation?
The expansion of $f(x_1)$ in terms of $w$ is
$$f(x_1 )=(γ^3 ) w^6+(3βγ^2 ) w^5+(aγ^2+3αγ^2+3β^2 γ) w^4
+(6αβγ+2aβγ+β^3 ) w^3+(2aαγ+aβ^2+3αβ^2+bγ+3α^2 γ) w^2
+(2aαβ+3α^2 β+bβ)w+(aα^2+αb+c+α^3 ),$$
however I do not see how it helps. Any elegant suggestions?
 [/INST] For the second part, it can be shown merely by calculation:
$$\begin{align}x_1+x_2+x_3&=(α+βw+γw^2 )+(α+ζβw+ζ^2γw^2 )+(α+ζ^2βw+ζγw^2)\\& =3α+βw(1+ζ+ζ^2 )+γw^2 (1+ζ+ζ^2 ),\end{align}$$
and by using $1+ζ+ζ^2=0$, as it is a root of $\Phi_3(x)=1+x+x^2=\frac{x^3-1}{x-1}$:
$$α=\frac{1}{3} (x_1 + x_2 + x_3).$$
In a similar fashion:
$$\begin{align}x_1+ζx_2+ζ^2x_3&=(α+βw+γw^2)+ζ(α+ζβw+ζ^2 γw^2)+ζ^2 (α+ζ^2 βw+ζγw^2 )\\&=α(1+ζ+ζ^2 )+βw(1+ζ^2+ζ^4 )+γw^2 (1+ζ^3+ζ^3 )\\&=3γw^2,\end{align}$$
and
$$\begin{align}x_1+x_2ζ^2+x_3ζ&=(α+βw+γw^2)+ζ^2 (α+ζβw+ζ^2 γw^2)+ζ(α+ζ^2 βw+ζγw^2)\\&=α(1+ζ+ζ^2 )+3βw+γw^2 (1+ζ^2+ζ^4)\\&=3\beta w.\end{align} $$
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