14. If $\sqrt[3]{x+5}-\sqrt[3]{x-5}=1$, find $x^{2}$.
14. 52
By factoring with difference of two cubes, we have
$$
\begin{array}{l}
(\sqrt[3]{x+5})^{3}-(\sqrt[3]{x-5})^{3} \\
=(\sqrt[3]{x+5}-\sqrt[3]{x-5})\left(\sqrt[3]{(x+5)^{2}}+\sqrt[3]{(x+5)(x-5)}+\sqrt[3]{(x-5)^{2}}\right)
\end{array}
$$
which can be simplified into
$$
(x+5)-(x-5)=\left(\sqrt[3]{(x+5)^{2}}+\sqrt[3]{x^{2}-25}+\sqrt[3]{(x-5)^{2}}\right)
$$
or
$$
\sqrt[3]{(x+5)^{2}}+\sqrt[3]{x^{2}-25}+\sqrt[3]{(x-5)^{2}}=10 .
$$

On the other hand, squaring the given equation, we get
$$
\sqrt[3]{(x+5)^{2}}-2 \sqrt[3]{x^{2}-25}+\sqrt[3]{(x-5)^{2}}=1 .
$$

Subtracting $(\star \star)$ from $(\star)$, we obtain
$$
3 \sqrt[3]{x^{2}-25}=9 \quad \text { or } \quad x^{2}=52
$$