1. Solve the inequality $\left(x^{2}-x+1\right)^{16 x^{3}-6 x} \leq\left(x^{2}-x+1\right)^{13 x^{2}+x^{3}}$.
Answer: $x \in\left(-\infty ;-\frac{1}{3}\right] \cup\{0\} \cup\left[1 ; \frac{6}{5}\right]$.

Solution. The base of the exponent is positive for all $x$, so the given inequality is equivalent to the following:

$$
\left.\left.\left(x^{2}-x\right)\right)\left(16 x^{3}-6 x\right)-\left(13 x^{2}+x^{3}\right)\right) \leq 0 \Leftrightarrow(x-1) x^{2}\left(15 x^{2}-13 x-6\right) \leq 0 \Leftrightarrow(x-1) x^{2}(5 x-6)(3 x+1) \leq 0,
$$

from which, using the method of intervals, we find that $x \in\left(-\infty ;-\frac{1}{3}\right] \cup\{0\} \cup\left[1 ; \frac{6}{5}\right]$.