Solve the following equation:

$$
\sqrt[3]{76+\sqrt{x}}+\sqrt[3]{76-\sqrt{x}}=8
$$
I. Solution: Let us introduce the following new unknowns:

$$
u=\sqrt[3]{76+\sqrt{x}}, \quad v=\sqrt[3]{76-\sqrt{x}}
$$

that is,

$$
\begin{aligned}
u^{3} & =76+\sqrt{x} \\
v^{3} & =76-\sqrt{x}
\end{aligned}
$$

The sum of (1) and (2)

$$
u^{3}+v^{3}=152
$$

From the original equation,

$$
u+v=8
$$

Dividing (3) by (4)

$$
u^{2}-u v+v^{2}=19
$$

The square of (4)

$$
u^{2}+2 u v+v^{2}=64
$$

The difference between (5) and (4)

$$
3 u v=45, \quad \text { or } \quad u v=15
$$

Thus,

$$
u^{3} v^{3}=(76+\sqrt{x})(76-\sqrt{x})=15^{3}
$$

which means

$$
76^{2}-x=15^{3}
$$

from which

$$
x=76^{2}-15^{3}=5776-3375=2401
$$

By substitution, we can verify that this value of $x$ satisfies our equation.

Simonfai László (Bp. II., Rákóczi g. II. o. t.)

II. Solution: Cubing both sides of the equation, the left side becomes:

$$
\begin{gathered}
76+\sqrt{x}+3(\sqrt[3]{76+\sqrt{x}})^{2} \cdot \sqrt[3]{76-\sqrt{x}}+3 \sqrt[3]{76+\sqrt{x}} \cdot(\sqrt[3]{76-\sqrt{x}})^{2}+76-\sqrt{x}= \\
=152+3 \sqrt[3]{76^{2}-x} \cdot(\sqrt[3]{76+\sqrt{x}}+\sqrt[3]{76-\sqrt{x}})=152-24 \sqrt[3]{76^{2}-x}
\end{gathered}
$$

Thus, we arrive at the equation

$$
\begin{gathered}
152+24 \sqrt[3]{76^{2}-x}=8^{3}=512 \\
\sqrt[3]{76^{2}-x}=\frac{512-152}{24}=15
\end{gathered}
$$

From here, cubing both sides and solving for $x$, we find that

$$
x=76^{2}-15^{3}=5776-3375=2401
$$

is the only possible solution to the equation. A simple calculation shows that this value indeed satisfies the equation.