[proofplan]
We work with $\mu = e^{2\pi i/3}$, a primitive cube root of unity, so $\mu^3 = 1$ and $1 + \mu + \mu^2 = 0$, i.e., $\mu^2 + \mu = -1$. The roots $\alpha_1, \alpha_2, \alpha_3$ of $f(t) = t^3 + pt + q$ satisfy Vieta's relations: $\alpha_1 + \alpha_2 + \alpha_3 = 0$, $\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3 = p$, $\alpha_1\alpha_2\alpha_3 = -q$. We compute the product $\beta\gamma$, then $\beta^3 + \gamma^3$, and finally assemble the resolvent quadratic for $w = \beta^3$ and $w = \gamma^3$.
[/proofplan]
[step:Compute the product $\beta\gamma$]
We expand $\beta\gamma$ by multiplying term-by-term:
\begin{align*}
\beta\gamma &= (\alpha_1 + \mu\alpha_2 + \mu^2\alpha_3)(\alpha_1 + \mu^2\alpha_2 + \mu\alpha_3).
\end{align*}
Distributing, we collect diagonal and cross terms. The squared terms give $\alpha_1^2 + \mu^3\alpha_2^2 + \mu^3\alpha_3^2 = \alpha_1^2 + \alpha_2^2 + \alpha_3^2$ since $\mu^3 = 1$. The cross terms are:
\begin{align*}
&(\mu^2 + \mu)\alpha_1\alpha_2 + (\mu + \mu^2)\alpha_1\alpha_3 + (\mu^2 + \mu^4)\alpha_2\alpha_3.
\end{align*}
Since $\mu^4 = \mu^3 \cdot \mu = \mu$, every parenthetical factor equals $\mu^2 + \mu = -1$. Therefore:
\begin{align*}
\beta\gamma &= \alpha_1^2 + \alpha_2^2 + \alpha_3^2 - \alpha_1\alpha_2 - \alpha_1\alpha_3 - \alpha_2\alpha_3.
\end{align*}
[guided]
Now use the identity $\alpha_1^2 + \alpha_2^2 + \alpha_3^2 = (\alpha_1 + \alpha_2 + \alpha_3)^2 - 2(\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3)$. Since $\alpha_1 + \alpha_2 + \alpha_3 = 0$ and $\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3 = p$, this gives $\alpha_1^2 + \alpha_2^2 + \alpha_3^2 = -2p$.
[/guided]
Substituting:
\begin{align*}
\beta\gamma &= -2p - p = -3p. \qquad \checkmark
\end{align*}
[/step]
[step:Compute $\beta^3 + \gamma^3$]
Rather than cubing directly, we exploit the cyclic structure. Observe that $\beta^3 = (\alpha_1 + \mu\alpha_2 + \mu^2\alpha_3)^3$. When we expand this cube, every monomial picks up a power of $\mu$. A monomial $\alpha_i \alpha_j \alpha_k$ (with $i,j,k$ drawn from positions in $\beta$) acquires the factor $\mu^{a}$ where $a$ is the sum of the exponents of $\mu$ attached to those positions. The key observation is that $\mu^a$ survives the sum $\beta^3 + \gamma^3$ only when $a \equiv 0 \pmod{3}$, because in $\gamma$ the roles of $\mu$ and $\mu^2$ are swapped, so $\gamma^3$ contributes $\mu^{b}$ with $b = 3 \cdot 3 - a \equiv -a$, and $\mu^a + \mu^{-a} = 2$ when $3 \mid a$ and $= -1$ otherwise.
[guided]
It is cleaner to use Newton's approach. Set $s = \beta + \gamma$ and $p_0 = \beta\gamma = -3p$. By direct computation, $\beta + \gamma = (\alpha_1 + \mu\alpha_2 + \mu^2\alpha_3) + (\alpha_1 + \mu^2\alpha_2 + \mu\alpha_3) = 2\alpha_1 + (\mu + \mu^2)(\alpha_2 + \alpha_3)$. Since $\mu + \mu^2 = -1$ and $\alpha_2 + \alpha_3 = -\alpha_1$, we get $s = 2\alpha_1 + \alpha_1 = 3\alpha_1$. So $\beta + \gamma = 3\alpha_1$, and similarly by cyclic relabeling we see these "Lagrange resolvents" depend on which root we single out. But we need $\beta^3 + \gamma^3$, not $s$.
[/guided]
We use the identity $\beta^3 + \gamma^3 = (\beta + \gamma)^3 - 3\beta\gamma(\beta + \gamma)$. From the guided note, $\beta + \gamma = 3\alpha_1$, so:
\begin{align*}
\beta^3 + \gamma^3 &= (3\alpha_1)^3 - 3(-3p)(3\alpha_1) \\
&= 27\alpha_1^3 + 27p\alpha_1.
\end{align*}
Since $\alpha_1$ is a root of $t^3 + pt + q = 0$, we have $\alpha_1^3 = -p\alpha_1 - q$. Substituting:
\begin{align*}
\beta^3 + \gamma^3 &= 27(-p\alpha_1 - q) + 27p\alpha_1 \\
&= -27p\alpha_1 - 27q + 27p\alpha_1 \\
&= -27q. \qquad \checkmark
\end{align*}
[/step]
[step:Form the resolvent quadratic]
Let $w_1 = \beta^3$ and $w_2 = \gamma^3$. From Steps 1 and 2 we know the elementary symmetric functions of $w_1, w_2$:
\begin{align*}
w_1 + w_2 &= \beta^3 + \gamma^3 = -27q, \\
w_1 w_2 &= (\beta\gamma)^3 = (-3p)^3 = -27p^3.
\end{align*}
[guided]
The quadratic with roots $w_1, w_2$ is $w^2 - (w_1 + w_2)w + w_1 w_2 = 0$. Substitute the values just computed.
[/guided]
Therefore $w_1$ and $w_2$ are the two roots of:
\begin{align*}
w^2 - (-27q)\,w + (-27p^3) &= 0, \\
w^2 + 27q\,w - 27p^3 &= 0. \qquad \blacksquare
\end{align*}
[/step]
