Problem 2. Calculate the real solutions of the equation:

$$
\sqrt[4]{97-X}+\sqrt[4]{X}=5
$$
Solution. If we let $a=\sqrt[4]{97-x}$ and $b=\sqrt[4]{x}$, the following is verified:

$$
\left.\begin{array}{l}
a+b=5 \\
a^{4}+b^{4}=97
\end{array}\right\}
$$

To solve this system, we proceed as follows:

$$
25=(a+b)^{2}=a^{2}+2 a b+b^{2}
$$

from which we have $a^{2}+b^{2}=25-2 a b$. On the other hand,

$$
625=(a+b)^{4}=a^{4}+b^{4}+4 a b\left(a^{2}+b^{2}\right)+6 a^{2} b^{2}=97+100 a b-2 a^{2} b^{2}
$$

from which we have $a^{2} b^{2}-50 a b+264=0$. Then $a b$ is a root of $z^{2}-50 z+264=0$; since the roots are 44 and 6, we study each of these cases.

If $a b=44$, then $a$ and $b$ are the solutions of the system $\left.\begin{array}{l}a+b=5 \\ a b=44\end{array}\right\}$, hence from the equation $z^{2}-5 z+44=0$; this does not have real roots. If $a b=6$, then $a$ and $b$ are roots of the equation $z^{2}-5 z+6=0$, which has the roots 3 and 2.

If $b=3$, then $x=81$, and if $b=2$, then $x=16$. These are the only real roots of the given equation.