24.12. Solve the equation $x^{5}-5 a x^{3}+5 a^{2} x-b=0$.
24.12. We will use the same method that was used in problem 24.2 to solve the cubic equation. Specifically, we will look for a solution in the form $x=\alpha+\beta$. Using the fact that $(\alpha+\beta)^{3}=\alpha^{3}+\beta^{3}+3 \alpha \beta(\alpha+\beta)$ and $(\alpha+\beta)^{5}=\alpha^{5}+\beta^{5}+5 \alpha \beta\left(\alpha^{3}+\beta^{3}\right)+$ $+10 \alpha^{2} \beta^{2}(\alpha+\beta)$, we can write the original equation in the form $\alpha^{5}+\beta^{5}+$ $+A\left(\alpha^{3}+\beta^{3}\right)+B(\alpha+\beta)-b=0$, where $A=5 \alpha \beta-5 a$ and $B=10 \alpha^{2} \beta^{2}-$ $-15 a \alpha \beta+5 a^{2}$. If $\alpha \beta=a$, then $A=0$ and $B=0$. Therefore, the solution of the original equation has been reduced to solving the system of equations $\alpha^{5}+\beta^{5}=b, \alpha \beta=a$. From this system, we find $\alpha^{5}, \beta^{5}=\frac{b}{2} \pm \sqrt{\frac{b^{2}}{4}-a^{5}}$.
Therefore,

$$
x=\sqrt[5]{\frac{b}{2}+\sqrt{\frac{b^{2}}{4}-a^{5}}}+\sqrt[5]{\frac{b}{2}-\sqrt{\frac{b^{2}}{4}-a^{5}}}
$$

The values of the roots are chosen such that their product is equal to $a$.