Determine all real numbers $x$ for which $\sqrt[3]{(2+x)^{2}}+3 \sqrt[3]{(2-x)^{2}}=4 \sqrt[3]{4-x^{2}}$.
We use the following facts about exponents:

- $\sqrt[3]{a}=a^{1 / 3}$ for every real number $a$
- Every real number can be squared
- The cube root of every number exists
- $\left(b^{c}\right)^{d}=b^{c d}=b^{d c}=\left(b^{d}\right)^{c}$ whenever $b^{c}$ and $b^{d}$ and $b^{c d}$ all exist
- $(m n)^{p}=m^{p} n^{p}$ whenever $m^{p}$ and $n^{p}$ and $(m n)^{p}$ all exist

Manipulating the given equation, we obtain

$$
\begin{aligned}
\sqrt[3]{(2+x)^{2}}+3 \sqrt[3]{(2-x)^{2}} & =4 \sqrt[3]{4-x^{2}} \\
\left((2+x)^{2}\right)^{1 / 3}+3\left((2-x)^{2}\right)^{1 / 3} & =4\left(4-x^{2}\right)^{1 / 3} \\
(2+x)^{2 / 3}+3(2-x)^{2 / 3} & =4((2+x)(2-x))^{1 / 3} \\
\left((2+x)^{1 / 3}\right)^{2}+3\left((2-x)^{1 / 3}\right)^{2} & =4(2+x)^{1 / 3}(2-x)^{1 / 3}
\end{aligned}
$$

Letting $u=(2+x)^{1 / 3}$ and $v=(2-x)^{1 / 3}$, this equation becomes

$$
\begin{aligned}
u^{2}+3 v^{2} & =4 u v \\
u^{2}-4 u v+3 v^{2} & =0 \\
(u-v)(u-3 v) & =0
\end{aligned}
$$

Therefore, $u=v$ or $u=3 v$.

If $u=v$, then $(2+x)^{1 / 3}=(2-x)^{1 / 3}$.

Cubing both sides gives $2+x=2-x$ and so $2 x=0$ or $x=0$.

If $u=3 v$, then $(2+x)^{1 / 3}=3(2-x)^{1 / 3}$.

Cubing both sides gives $2+x=3^{3}(2-x)$ and so $2+x=54-27 x$ or $28 x=52$ or $x=\frac{52}{28}=\frac{13}{7}$.

We verify these answers in the original equation.

If $x=0$, the left side of the equation becomes

$$
\sqrt[3]{(2+0)^{2}}+3 \sqrt[3]{(2-0)^{2}}=\sqrt[3]{4}+3 \sqrt[3]{4}=4 \sqrt[3]{4}=4 \sqrt[3]{4-0^{2}}
$$

which equals the right side of the equation, and so $x=0$ is a solution.

If $x=\frac{13}{7}$, the right side of the equation equals

$$
4 \sqrt[3]{4-\left(\frac{13}{7}\right)^{2}}=4 \sqrt[3]{4-\frac{169}{49}}=4 \sqrt[3]{\frac{196-169}{49}}=4 \sqrt[3]{\frac{27}{49}}=4 \frac{3}{\sqrt[3]{49}}=\frac{12}{\sqrt[3]{49}}
$$

and the left side of the equation equals

$$
\sqrt[3]{\left(2+\frac{13}{7}\right)^{2}}+3 \sqrt[3]{\left(2-\frac{13}{7}\right)^{2}}=\sqrt[3]{\left(\frac{27}{7}\right)^{2}}+3 \sqrt[3]{\left(\frac{1}{7}\right)^{2}}=\sqrt[3]{\frac{729}{49}}+3 \sqrt[3]{\frac{1}{49}}=\frac{9}{\sqrt[3]{49}}+3 \frac{1}{\sqrt[3]{49}}=\frac{12}{\sqrt[3]{49}}
$$

(We used the fact that $729=9^{3}$.) Therefore, the left side equals the right side, as required, and so $x=\frac{13}{7}$ is also a solution.

Thus, the solutions to the original equation are $x=0$ and $x=\frac{13}{7}$.