Irrational Equations

[ Variable Substitution (etc.).

Let $a$ be a given real number, $n$ be a natural number, $n>1$.

Find all such $x$ that the sum of the $n$-th roots of the numbers $x^{n}-a^{n}$ and $2 a^{n}-x^{n}$ equals the number $a$.

#
Let's first consider the special case $a=0$. We get: $\sqrt[n]{x^{n}}+\sqrt[n]{-x^{n}}=0$.

If $n$ is even, the only solution will be $x=0$ (otherwise one of the roots is undefined). If $n$ is odd, any $x$ will be a solution.

Now let $a \neq 0$. Denote $\sqrt[n]{x^{n}-a^{n}}$ by $y$, and $a-y$ by $z$, we get the system: $y+z=a, y^{n}+z^{n}=a^{n}$. Let $y=a, z=0$ or $y=0, z=a$. In these cases, both equations are satisfied. Let's see what solutions these give. We compute $x: x^{n}=2 a^{n}$ or $x^{n}=a^{n}$.

These equations are satisfied when $x=a \sqrt[n]{2}$ and $x=a$, if $n$ is odd, and when $x= \pm a \sqrt[n]{2}$ and $x= \pm a$, if $n$ is even (in this case $a \geq 0$). Substituting these values of $x$ into the original equation, we get the same result in both cases: $\sqrt[n]{a^{n}}=a$.

Now let's prove that when $a \neq 0$, either $y$ or $z$ is zero. It is sufficient to consider the case $a>0, y \geq z$.

Indeed, if $(y, z, a)$ is a solution, then $(-y,-z,-a)$ is also a solution. Furthermore, if we swap the values of $y$ and $z$ in a solution, we get another solution. Consider two cases.

a) $z>0$. But then the equality $y^{n}+z^{n}=(y+z)^{n}$ is impossible.

b) $z=-wa$, from which $y^{n}+w^{n}>y^{n}>a^{n}$.

For odd $n$, the system becomes

$$
\begin{aligned}
& y-w=a \\
& y^{n}-w^{n}=a^{n} \\
& a+w=y \\
& a^{n}+w^{n}=y^{n}
\end{aligned}
$$

or

which is analogous to case a).

## Answer

If $n$ is odd, then for $a \neq 0 \quad x_{1}=a \sqrt[n]{2}, x_{2}=a ;$ for $a=0 \quad x$-any.

![](https://cdn.mathpix.com/cropped/2024_05_06_0ff1ee02df212bd75258g-44.jpg?height=71&width=1755&top_left_y=1668&top_left_x=0)