24.10. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of the equation $x^{4} + a x^{2} + b x + c = 0$. Let $\alpha = -\left(x_{1} + x_{2}\right)\left(x_{3} + x_{4}\right)$, $\beta = -\left(x_{1} + x_{3}\right) \times$ $\times\left(x_{2} + x_{4}\right)$, and $\gamma = -\left(x_{1} + x_{4}\right)\left(x_{2} + x_{3}\right)$.

a) Express $x_{1}, x_{2}, x_{3}$, and $x_{4}$ in terms of $\sqrt{\alpha}, \sqrt{\beta}$, and $\sqrt{\gamma}$.

b) Express the coefficients of the polynomial $(y-\alpha)(y-\beta)(y-\gamma)$ in terms of $a, b, c$.

c) Reduce the solution of the equation $x^{4} + a x^{2} + b x + c = 0$ to the solution of a cubic equation.

## 24.4. Other Equations Solvable in Radicals
24.10. a) By the condition $x_{1}+x_{2}=-\left(x_{3}+x_{4}\right)$, therefore $x_{1}+x_{2}=\sqrt{\alpha}$. Similarly, $x_{1}+x_{3}=\sqrt{\beta}$ and $x_{1}+x_{4}=\sqrt{\gamma}$. Consequently,

$$
\begin{aligned}
& 2 x_{1}=\sqrt{\alpha}+\sqrt{\beta}+\sqrt{\gamma} \\
& 2 x_{2}=\sqrt{\alpha}-\sqrt{\beta}-\sqrt{\gamma} \\
& 2 x_{3}=-\sqrt{\alpha}+\sqrt{\beta}-\sqrt{\gamma} \\
& 2 x_{4}=-\sqrt{\alpha}-\sqrt{\beta}+\sqrt{\gamma}
\end{aligned}
$$

b) It is clear that $\alpha+\beta+\gamma=-2\left(x_{1} x_{2}+x_{2} x_{3}+\ldots\right)=-2 a$. Further, $\alpha \beta+\gamma \alpha+\beta \gamma=\left(x_{1}^{2} x_{2}^{2}+\ldots\right)+3\left(x_{1}^{2} x_{2} x_{3}+\ldots\right)+6 x_{1} x_{2} x_{3} x_{4}$. Considering that $a^{2}=\left(x_{1} x_{2}+\ldots\right)^{2}=\left(x_{1}^{2} x_{2}^{2}+\ldots\right)+2\left(x_{1}^{2} x_{2} x_{3}+\ldots\right)+6 x_{1} x_{2} x_{3} x_{4}$ and

$$
0=\left(x_{1}+\ldots\right)\left(x_{1} x_{2} x_{3}+\ldots\right)=\left(x_{1}^{2} x_{2} x_{3}+\ldots\right)+4 x_{1} x_{2} x_{3} x_{4}
$$

we get $\alpha \beta+\gamma \alpha+\beta \gamma=a^{2}-4 c$. Finally, $-\alpha \beta \gamma=\left(x_{1}^{3} x_{2}^{2} x_{3}+\ldots\right)+$ $+2\left(x_{1}^{3} x_{2} x_{3} x_{4}+\ldots\right)+2\left(x_{1}^{2} x_{2}^{2} x_{3}^{2}+\ldots\right)+4\left(x_{1}^{2} x_{2}^{2} x_{3} x_{4}+\ldots\right)$. Considering that

$$
\begin{aligned}
& 0=\left(x_{1}+\ldots\right)\left(x_{1} x_{2}+\ldots\right)\left(x_{1} x_{2} x_{3}+\ldots\right)= \\
& =\left(x_{1}^{3} x_{2}^{2} x_{3}+\ldots\right)+3\left(x_{1}^{3} x_{2} x_{3} x_{4}+\ldots\right)+3\left(x_{1}^{2} x_{2}^{2} x_{3}^{2}+\ldots\right)+8\left(x_{1}^{2} x_{2}^{2} x_{3} x_{4}+\ldots\right) \\
& \quad 0=\left(x_{1}+\ldots\right)^{2} x_{1} x_{2} x_{3} x_{4}=\left(x_{1}^{3} x_{2} x_{3} x_{4}+\ldots\right)+2\left(x_{1}^{2} x_{2}^{2} x_{3} x_{4}+\ldots\right)
\end{aligned}
$$

and $b^{2}=\left(x_{1} x_{2} x_{3}+\ldots\right)^{2}=\left(x_{1}^{2} x_{2}^{2} x_{3}^{2}+\ldots\right)+2\left(x_{1}^{2} x_{2}^{2} x_{3} x_{4}+\ldots\right)$, we get $\alpha \beta \gamma=b^{2}$.

c) According to problem a) it is sufficient to find $\alpha, \beta$ and $\gamma$. According to problem b) the numbers $\alpha, \beta$ and $\gamma$ are the roots of the cubic equation $y^{3}+2 a y^{2}+\left(a^{2}-4 c\right) y-b^{2}=0$.