1. (12 points) Solve the equation

$$
\sqrt[3]{(9-x)^{2}}-\sqrt[3]{(9-x)(7+x)}+\sqrt[3]{(7+x)^{2}}=4
$$
Answer: 1.

Solution. Let $a=9-x, b=7+x$. Note that $a+b=16$. We obtain the system of equations

$$
\left\{\begin{array}{c}
\sqrt[3]{a^{2}}-\sqrt[3]{a b}+\sqrt[3]{b^{2}}=4 \\
a+b=16
\end{array}\right.
$$

Multiply the first equation by $\sqrt[3]{a}+\sqrt[3]{b}$, using the formula for the difference of cubes, we get $\sqrt[3]{a}+\sqrt[3]{b}=4$. Square this equality, and subtract the first equation of the system from the result, we get $\sqrt[3]{a b}=4$.

The system $\left\{\begin{array}{c}\sqrt[3]{a b}=4, \\ a+b=16\end{array}\right.$ reduces to a quadratic equation. Solving it, we get $a=8 ; x=1$. Let's check the solution.

Evaluation criteria. Full justified solution - 12 points. Correct answer obtained without justification (guessed) - 3 points.