Problem 5. Calculate the real solutions of the equation:

$$
\sqrt[3]{1729-X}+\sqrt[3]{X}=19
$$
Solution. If we let $a=\sqrt[3]{1729-x}$ and $b=\sqrt[3]{x}$, then $a$ and $b$ are roots of the system $\left.\begin{array}{l}a+b=19 \\ a^{3}+b^{3}=1729\end{array}\right\}$. To solve this system, we proceed as follows:

$$
19^{3}=(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)=1729+3 \times 19 a b
$$

from which we have $a b=90$. Thus, $a$ and $b$ are the roots of the system $\left.\begin{array}{l}a+b=19 \\ a b=90\end{array}\right\}$, and therefore of the equation $z^{2}-19 z+90=0$. Since the roots of this equation are 10 and 9, it follows that $x=10^{3}=1000$ and $x=9^{3}=729$ are the only real roots of the given equation.