Ionin Yu. I.

Let $p$ be an arbitrary real number. Find all such $x$ that the sum of the cube roots of the numbers $1-x$ and $1+x$ equals $p$.
Let $u^{3}=1-x, v^{3}=1+x$. From the equation $u+v=p$ we get $v^{3}+v^{3}+3 u v(u+v)=p^{3}$, hence $2+3 u v p=p^{3}$, that is $u^{3} v^{3}=\left(\frac{p^{3}-2}{3 p}\right)^{3}$. Now $u^{3}$ and $v^{3}$ are found as the roots of the quadratic equation $t^{2}-2 t+\left(\frac{p^{3}-2}{3 p}\right)^{3}=0$.

In the end, we get $x= \pm \sqrt{1-\left(\frac{p^{3}-2}{3 p}\right)^{3}}$.

Since $1-\frac{p^{3}-2}{3 p}=\frac{-p^{3}+3 p-2}{3 p}=\frac{-(p-1)^{2}(p+2)}{3 p}$, the obtained answer makes sense for $p=$ -1 and $0<p \leq 2$.

However, extraneous roots could have arisen from our transformations. Indeed, the equation $u^{3}+v^{3}-$ $p^{3}+3 u v p=0$ can be written as $(u+v)^{3}-p^{3}-3 u v(u+v-p)=0 \Leftrightarrow(u+v-p)\left(u^{2}+v^{2}+p^{2}+u p+p v-u v\right)=$ 0.

The left side is zero when at least one of the brackets is zero:

1) $u+v-p=0$,
2) $u^{2}+v^{2}+p^{2}-(u p+p v-u v)=0$.

The first equality is our original equation. All extraneous solutions are related to the second, and we need to determine when it holds. Multiplying it by 2, we get: $(u+p)^{2}+(v+p)^{2}+(u-v)^{2}=0$. This equality holds only when $u=v=-p$, that is, when $x=0, p=-1$.

And indeed $(1+x)^{1 / 3}+(1-x)^{1 / 3} \neq-1$ when $x=0$.

## Answer

For $0<p \leq 2 \quad x= \pm \sqrt{1-\left(\frac{p^{3}-2}{3 p}\right)^{3}}$, for other values of $p$ such $x$ does not exist.