[proofplan] We know $\alpha_1+\alpha_2+\alpha_3+\alpha_4=0$, so $\beta+\gamma+\lambda = 3\alpha_1+(\alpha_2+\alpha_3+\alpha_4)=3\alpha_1-\alpha_1=2\alpha_1$, and more critically $\beta\gamma\lambda$ and the pairwise products can be expressed through the coefficients $b,c,d$ of the reduced quartic. We compute $e_1=\beta^2+\gamma^2+\lambda^2$, $e_2=\beta^2\gamma^2+\beta^2\lambda^2+\gamma^2\lambda^2$, and $e_3=\beta^2\gamma^2\lambda^2$ by first expressing the elementary symmetric polynomials of $\beta,\gamma,\lambda$ in terms of $b,c,d$, then lifting to $\beta^2,\gamma^2,\lambda^2$ via Newton-type identities. The results $e_1=-2b$, $e_2=b^2-4d$, $e_3=c^2$ confirm that $g(t)=t^3+2bt^2+(b^2-4d)t-c^2$. [/proofplan]

[step:Elementary symmetric polynomials of $\beta,\gamma,\lambda$]Write Vieta's relations for $f(t)=t^4+bt^2+ct+d$ with roots $\alpha_1,\alpha_2,\alpha_3,\alpha_4$: \begin{align*} \alpha_1+\alpha_2+\alpha_3+\alpha_4 &= 0, \\ \sum_{i<j}\alpha_i\alpha_j &= b, \\ \sum_{i<j<k}\alpha_i\alpha_j\alpha_k &= -c, \\ \alpha_1\alpha_2\alpha_3\alpha_4 &= d. \end{align*} Since the roots sum to zero, $\alpha_3+\alpha_4=-\beta$, $\alpha_2+\alpha_4=-\gamma$, $\alpha_2+\alpha_3=-\lambda$. Therefore \begin{align*} \beta+\gamma+\lambda &= (\alpha_1+\alpha_2)+(\alpha_1+\alpha_3)+(\alpha_1+\alpha_4) = 3\alpha_1+(\alpha_2+\alpha_3+\alpha_4) = 3\alpha_1-\alpha_1 = 2\alpha_1. \end{align*} By symmetry each of $2\alpha_1,2\alpha_2,2\alpha_3,2\alpha_4$ arises from a different labelling, but we will not need $\beta+\gamma+\lambda$ individually. Instead, compute: **Sum of products in pairs.** Expand $\beta\gamma+\beta\lambda+\gamma\lambda$: \begin{align*} \beta\gamma &= (\alpha_1+\alpha_2)(\alpha_1+\alpha_3) = \alpha_1^2+\alpha_1\alpha_3+\alpha_1\alpha_2+\alpha_2\alpha_3, \\ \beta\lambda &= (\alpha_1+\alpha_2)(\alpha_1+\alpha_4) = \alpha_1^2+\alpha_1\alpha_4+\alpha_1\alpha_2+\alpha_2\alpha_4, \\ \gamma\lambda &= (\alpha_1+\alpha_3)(\alpha_1+\alpha_4) = \alpha_1^2+\alpha_1\alpha_4+\alpha_1\alpha_3+\alpha_3\alpha_4. \end{align*} Summing these three expressions: \begin{align*} \beta\gamma+\beta\lambda+\gamma\lambda &= 3\alpha_1^2 + 2\alpha_1(\alpha_2+\alpha_3+\alpha_4) + (\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4). \end{align*} \begin{align*} \beta\gamma+\beta\lambda+\gamma\lambda &= 3\alpha_1^2 + 2\alpha_1(-\alpha_1) + (b+\alpha_1^2) = 3\alpha_1^2-2\alpha_1^2+b+\alpha_1^2 = 2\alpha_1^2+b. \end{align*} We will also need $(\beta+\gamma+\lambda)^2 = 4\alpha_1^2$, hence \begin{align*} \beta^2+\gamma^2+\lambda^2 &= (\beta+\gamma+\lambda)^2 - 2(\beta\gamma+\beta\lambda+\gamma\lambda) = 4\alpha_1^2-2(2\alpha_1^2+b) = -2b. \end{align*} This is independent of $\alpha_1$, confirming it is a symmetric function of the coefficients. **Product $\beta\gamma\lambda$.** We have \begin{align*} \beta\gamma\lambda &= (\alpha_1+\alpha_2)(\alpha_1+\alpha_3)(\alpha_1+\alpha_4). \end{align*} Meanwhile $f(-\alpha_1)=\alpha_1^4+b\alpha_1^2-c\alpha_1+d$. Since $\alpha_1$ is a root, $\alpha_1^4=-b\alpha_1^2-c\alpha_1-d$, so \begin{align*} f(-\alpha_1) &= (-b\alpha_1^2-c\alpha_1-d)+b\alpha_1^2-c\alpha_1+d = -2c\alpha_1. \end{align*} Therefore $2\alpha_1\,\beta\gamma\lambda = -2c\alpha_1$, giving $\beta\gamma\lambda = -c$ (for $\alpha_1\neq 0$; the identity extends by continuity when $\alpha_1=0$, which forces $c=0$).[/step]

[guided]Use $\alpha_2+\alpha_3+\alpha_4=-\alpha_1$ and note that $\sum_{i<j}\alpha_i\alpha_j = b$ splits as $\alpha_1(\alpha_2+\alpha_3+\alpha_4)+(\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4)=b$, so $\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4 = b+\alpha_1^2$.[/guided]

[step:Symmetric functions of $\beta^2,\gamma^2,\lambda^2$ and identification of $g$]We now assemble $e_1,e_2,e_3$ for the squares $\beta^2,\gamma^2,\lambda^2$. **First symmetric function.** \begin{align*} e_1 = \beta^2+\gamma^2+\lambda^2 = -2b. \end{align*} **Third symmetric function.** This is immediate from Step 1: \begin{align*} e_3 = \beta^2\gamma^2\lambda^2 = (\beta\gamma\lambda)^2 = (-c)^2 = c^2. \end{align*} **Second symmetric function.** We compute $e_2 = \beta^2\gamma^2+\beta^2\lambda^2+\gamma^2\lambda^2 = (\beta\gamma+\beta\lambda+\gamma\lambda)^2 - 2\beta\gamma\lambda(\beta+\gamma+\lambda)$. Substituting the values from Step 1: \begin{align*} e_2 &= (2\alpha_1^2+b)^2 - 2(-c)(2\alpha_1) \\ &= 4\alpha_1^4 + 4b\alpha_1^2 + b^2 + 4c\alpha_1. \end{align*} Using $\alpha_1^4 = -b\alpha_1^2 - c\alpha_1 - d$ once more: \begin{align*} e_2 &= 4(-b\alpha_1^2-c\alpha_1-d)+4b\alpha_1^2+b^2+4c\alpha_1 \\ &= -4b\alpha_1^2-4c\alpha_1-4d+4b\alpha_1^2+b^2+4c\alpha_1 \\ &= b^2-4d. \end{align*} All dependence on $\alpha_1$ has cancelled, as expected. **Constructing $g$.** The monic cubic with roots $\beta^2,\gamma^2,\lambda^2$ is \begin{align*} g(t) &= t^3 - e_1\,t^2 + e_2\,t - e_3 \\ &= t^3 -(-2b)\,t^2 + (b^2-4d)\,t - c^2 \\ &= t^3 + 2bt^2 + (b^2-4d)t - c^2. \qquad\blacksquare \end{align*}[/step]

[guided]This is the standard identity $\sum_{i<j}x_i^2x_j^2 = \bigl(\sum_{i<j}x_ix_j\bigr)^2 - 2x_1x_2x_3\bigl(\sum x_i\bigr)$, applied with $x_i\in\{\beta,\gamma,\lambda\}$.[/guided]