24.13. What seventh-degree equations can be solved in the same way as was used in solving problem $24.12?$
24.13. Answer: $x^{7}-7 a x^{5}+14 a^{2} x^{3}-7 a^{3} x-b=0$.

Let $\alpha \beta=a$. Then

$$
\begin{array}{lr}
(\alpha+\beta)^{7}=\alpha^{7}+\beta^{7}+7 a\left(\alpha^{5}+\beta^{5}\right)+21 a^{2}\left(\alpha^{3}+\beta^{3}\right)+35 a^{3}(\alpha+\beta) \\
(\alpha+\beta)^{5}= & \alpha^{5}+\beta^{5}+5 a\left(\alpha^{3}+\beta^{3}\right)+10 a^{2}(\alpha+\beta) \\
(\alpha+\beta)^{3}= & \alpha^{3}+\beta^{3}+3 a(\alpha+\beta)
\end{array}
$$

To leave only $\alpha^{7}+\beta^{7}$, we need to consider the expression

$$
(\alpha+\beta)^{7}-7 a(\alpha+\beta)^{5}+14 a^{2}(\alpha+\beta)^{3}-7 a^{3}(\alpha+\beta)
$$

Thus, solving the equation $x^{7}-7 a x^{5}+14 a^{2} x^{3}-7 a^{3} x=b$ reduces to solving the system of equations $\alpha \beta=a, \alpha^{7}+\beta^{7}=b$.