## Task A-2.3.

Determine all real numbers $x$ for which

$$
3 \sqrt[3]{(14-x)^{2}}-2 \sqrt[3]{(14+x)^{2}}=5 \sqrt[3]{x^{2}-196}
$$
## Solution.

Notice that $x=14$ and $x=-14$ are not solutions to the equation.

Let $a=\sqrt[3]{14-x}, b=\sqrt[3]{14+x}$. Then $a b=\sqrt[3]{196-x^{2}}$, and the original equation can be written as $3 a^{2}-2 b^{2}=-5 a b$.

If $b$ were zero, then $x$ would be -14, which we have verified is not a solution to the equation. Therefore, $b \neq 0$.

Dividing by $b^{2}$, we get

$$
3 \cdot \frac{a^{2}}{b^{2}}-2=-5 \cdot \frac{a}{b}
$$

Substituting $t=\frac{a}{b}$ and rearranging, we get $3 t^{2}+5 t-2=0$.

The solutions to this quadratic equation are $t_{1}=\frac{1}{3}, t_{2}=-2$.

For $\frac{\sqrt[3]{14-x}}{\sqrt[3]{14+x}}=\frac{1}{3}$, we get $\frac{14-x}{14+x}=\frac{1}{27}$, which gives $x_{1}=13$.

Similarly, for $\frac{\sqrt[3]{14-x}}{\sqrt[3]{14+x}}=-2$, we get the second solution $x_{2}=-18$.

Note: Instead of dividing by $b^{2}$ and introducing the substitution $t$, the equation $3 a^{2}-2 b^{2}+5 a b=0$ can be written as $(3 a-b)(a+2 b)=0$, which is worth 3 points. After that, the case $3 a=b$ and obtaining the solution $x=13$ is worth 2 points, and the case $a=-2 b$ and $x=-18$ is worth another 2 points.