[proofplan]
We invert the linear system that defines the Lagrange resolvers. The three equations — Vieta's relation $\alpha_1+\alpha_2+\alpha_3=0$ and the definitions of $\beta,\gamma$ — form a system whose coefficient matrix is Vandermonde in $1,\mu,\mu^2$. We solve for each $\alpha_i$ using the key identity $1+\mu+\mu^2=0$, then verify the expressions are correct.
[/proofplan]
[step:Set Up the Linear System]
Since $t^3+pt+q$ has no $t^2$ term, Vieta's formulas give $\alpha_1+\alpha_2+\alpha_3=0$. Together with the definitions of the Lagrange resolvers, we have three equations in three unknowns:
\begin{align*}
\alpha_1 + \alpha_2 + \alpha_3 &= 0, \\
\alpha_1 + \mu\,\alpha_2 + \mu^2\alpha_3 &= \beta, \\
\alpha_1 + \mu^2\alpha_2 + \mu\,\alpha_3 &= \gamma.
\end{align*}
In matrix form this is
\begin{align*}
\begin{pmatrix} 1 & 1 & 1 \\ 1 & \mu & \mu^2 \\ 1 & \mu^2 & \mu \end{pmatrix} \begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{pmatrix} = \begin{pmatrix} 0 \\ \beta \\ \gamma \end{pmatrix}.
\end{align*}
[guided]
The coefficient matrix is a $3\times 3$ Vandermonde matrix $V$ with nodes $1,\mu,\mu^2$. A Vandermonde matrix with nodes $x_0,x_1,x_2$ has determinant $\prod_{0\le i<j\le 2}(x_j-x_i)$. Here that product is $(\mu-1)(\mu^2-1)(\mu^2-\mu)$. Since $\mu$ is a primitive cube root of unity, all three factors are nonzero — no two of $1,\mu,\mu^2$ coincide — so $\det V\neq 0$ and the system has a unique solution.
[/guided]
[/step]
[step:Invert the System to Recover Each Root]
We solve by adding suitable multiples of the three rows. Write the equations as (I), (II), (III).
**Finding $\alpha_1$.** Add all three equations:
\begin{align*}
3\alpha_1 + (1+\mu+\mu^2)\alpha_2 + (1+\mu^2+\mu)\alpha_3 &= 0+\beta+\gamma.
\end{align*}
Since $1+\mu+\mu^2=0$, the coefficients of $\alpha_2$ and $\alpha_3$ vanish, giving $3\alpha_1=\beta+\gamma$, hence
\begin{align*}
\alpha_1 = \frac{\beta+\gamma}{3}.
\end{align*}
**Finding $\alpha_2$.** Compute $(\text{I})+\mu^2(\text{II})+\mu(\text{III})$:
\begin{align*}
(1+\mu^2+\mu)\alpha_1 + (1+\mu^3+\mu^3)\alpha_2 + (1+\mu^4+\mu^3)\alpha_3 &= 0+\mu^2\beta+\mu\gamma.
\end{align*}
The coefficient of $\alpha_1$ is $1+\mu+\mu^2=0$. For $\alpha_2$ we use $\mu^3=1$, so the coefficient is $1+1+1=3$. For $\alpha_3$ the coefficient is $1+\mu^4+\mu^3=1+\mu+1=(2+\mu)$; but $\mu^4=\mu^3\cdot\mu=\mu$ and $\mu^3=1$, so it is $1+\mu+1=2+\mu$. We need to be more careful: the coefficient of $\alpha_3$ in each row is $1,\mu^2,\mu$, so the combination gives
\begin{align*}
1\cdot 1+\mu^2\cdot\mu^2+\mu\cdot\mu = 1+\mu^4+\mu^2 = 1+\mu+\mu^2 = 0,
\end{align*}
where we used $\mu^4=\mu$. Therefore $3\alpha_2=\mu^2\beta+\mu\gamma$, hence
\begin{align*}
\alpha_2 = \frac{\mu^2\beta+\mu\gamma}{3}.
\end{align*}
**Finding $\alpha_3$.** Compute $(\text{I})+\mu(\text{II})+\mu^2(\text{III})$. The coefficient of $\alpha_1$ is $1+\mu+\mu^2=0$. The coefficient of $\alpha_2$ is
\begin{align*}
1\cdot 1+\mu\cdot\mu+\mu^2\cdot\mu^2 = 1+\mu^2+\mu^4 = 1+\mu^2+\mu = 0.
\end{align*}
The coefficient of $\alpha_3$ is
\begin{align*}
1\cdot 1+\mu\cdot\mu^2+\mu^2\cdot\mu = 1+\mu^3+\mu^3 = 1+1+1 = 3.
\end{align*}
The right-hand side is $0+\mu\beta+\mu^2\gamma$, so
\begin{align*}
\alpha_3 = \frac{\mu\beta+\mu^2\gamma}{3}.
\end{align*}
[guided]
The pattern behind all three calculations is the same: we are computing the inverse of the Vandermonde matrix $V$. Since the nodes are the three cube roots of unity, $V/\sqrt{3}$ is (up to scaling) the discrete Fourier transform matrix of order 3, and its inverse is $\overline{V}/3$. The three linear combinations $(\text{I})+\omega^k(\text{II})+\omega^{2k}(\text{III})$ for $k=0,1,2$ (with $\omega=\mu^2=\bar\mu$) are exactly the rows of this inverse, and the identity $1+\mu+\mu^2=0$ is what kills the off-diagonal terms each time.
[/guided]
[/step]
[step:Verify the Expressions Are Consistent]
As a check, substitute the claimed formulas back into the definition of $\beta$:
\begin{align*}
\alpha_1+\mu\alpha_2+\mu^2\alpha_3 &= \frac{1}{3}\bigl[(\beta+\gamma)+\mu(\mu^2\beta+\mu\gamma)+\mu^2(\mu\beta+\mu^2\gamma)\bigr] \\
&= \frac{1}{3}\bigl[\beta+\gamma+\mu^3\beta+\mu^2\gamma+\mu^3\beta+\mu^4\gamma\bigr] \\
&= \frac{1}{3}\bigl[\beta+\gamma+\beta+\mu^2\gamma+\beta+\mu\gamma\bigr] \\
&= \frac{3\beta}{3}+\frac{(1+\mu+\mu^2)\gamma}{3} = \beta+0 = \beta,
\end{align*}
using $\mu^3=1$, $\mu^4=\mu$, and $1+\mu+\mu^2=0$. An identical calculation confirms $\alpha_1+\mu^2\alpha_2+\mu\alpha_3=\gamma$, and summing the three $\alpha_i$ gives $(1+\mu^2+\mu)(\beta+\gamma)/3=0$, consistent with Vieta's relation. $\blacksquare$
[/step]
