58. Find all real $\alpha$ for which the system of equations

$$
\frac{a^{3}}{b+c+\alpha}=\frac{b^{3}}{c+a+\alpha}=\frac{c^{3}}{a+b+\alpha}
$$

has a solution in distinct real $a, b, c$ from $[-1 ; 1]$.
Answer: $\alpha \in(-1 / 3,1 / 3) \backslash\{0\}$. Let $\alpha+a+b+c=s, a^{3} /(b+c+\alpha)=p$. Clearly, $p \neq 0$, otherwise it would be $a=b=c=0$. So $a b c \neq 0$. The numbers $a, b, c$ are the roots of the polynomial

$$
f(x)=x^{3}+p x-p s=0
$$

By Vieta's theorem $a+b+c=0$, so $\alpha=s=(p s) / p=a b c /(a b+b c+a c), \alpha \neq 0$ and $\alpha^{-1}=1 / a+1 / b+1 / c$. If, for example, $a, b>0, c=-a-b<0$, then $a+b \leqslant 1$ and $\alpha^{-1}$ can take any values that $1 / a+1 / b-1 /(a+b)$ takes under these conditions (it is sufficient to take $p=a b+b c+a c$, since Vieta's theorem works both ways). Writing this expression as $1 / a+a /\left(b a+b^{2}\right)$ we see that it decreases with $b$, so it takes its minimum value when $b=1-a$, when it equals $1 / a b-1=1 /\left(a-a^{2}\right)-1 \geqslant 3$, equality at $a=b=1 / 2$. But equal values of $a$ and $b$ cannot be taken, so $\alpha^{-1}$ takes all values greater than 3. Similarly, when $a, b<0$, we get all values less than 3.