Solve the following system of equations:

$$
\sqrt[3]{x-3}+\sqrt[3]{y+4}=11, \quad x+y=340
$$
By introducing the new unknowns $u$ and $v$ in place of the two radical expressions,

$$
x=u^{3}+3, \quad y=v^{3}-4
$$

the system of equations becomes:

$$
\begin{gathered}
u+v=11 \\
u^{3}+v^{3}=341
\end{gathered}
$$

Writing the left side of (3) as $(u+v)\left(u^{2}-u v+v^{2}\right)$ and considering (2), we obtain a quadratic equation:

$$
u^{2}-u v+v^{2}=31
$$

Subtracting (4) from the square of (2), we get $3 u v=90, u v=30$. From this and (2), $u$ and $v$ are the roots of the following equation:

$$
t^{2}-11 t+30=0
$$

The roots of this equation are 5 and 6, so $u_{1}=5, v_{1}=6 ; u_{2}=6, v_{2}=5$, and finally, based on (1),

$$
x_{1}=128, \quad y_{1}=212 ; \quad x_{2}=219, \quad y_{2}=121
$$

László Papp (Jászberény, Lehel Vezér Gymn., II. o. t.)

Remark. Solutions that arrived at the result through a quadratic equation written for $\sqrt[3]{x-3}$ or $\sqrt[3]{y+4}$ differ from the above only in notation.

Substituting $v$ from (2) into (3) would yield an equation for $u$ that is identical to (5). The path followed more clearly shows the symmetry of $u$ and $v$, and why the degree decreases from 3 to 2.