391. Solve the system:

$$
\begin{aligned}
& x+y+z=26 \\
& x-y=4 \\
& x-z=6
\end{aligned}
$$
391. MacLaurin's solution. He adds all three equations term by term and gets \(3x = 36\), from which \(x = 12\), then from the 2nd \(y = 8\) and from the 3rd \(z = 6\).

\[
\begin{aligned}
& \text{392.} \quad \frac{\sqrt[3]{20}}{\sqrt[3]{4}-\sqrt[3]{2}}=\frac{\sqrt[3]{20}}{\sqrt[3]{4}-\sqrt[3]{2}} \cdot \frac{\sqrt[3]{16}+2+1_{1}^{\prime}}{\sqrt[3]{16}+2+\sqrt[3]{4}}= \\
& =\frac{\sqrt[3]{20}(2 \sqrt[3]{2}+2+\sqrt[3]{4})}{(\sqrt[3]{4}-\sqrt[3]{2})(2 \sqrt[3]{2}+2+\sqrt[3]{4})}=\frac{2 \sqrt[3]{40}+2 \sqrt[3]{20}+\sqrt[3]{80}}{2}= \\
& =2 \sqrt[3]{5}+\sqrt[3]{20}+\sqrt[3]{10}
\end{aligned}
\]