2.3. Find all values of $x$ for which there exists at least one $a, -1 \leq a \leq 2$, such that the inequality holds: $(2-a) x^{3}+(1-2 a) x^{2}-6 x+\left(5+4 a-a^{2}\right)<0$.
2.3. $f(a)=a^{2}+a\left(x^{3}+2 x^{2}-4\right)-\left(2 x^{3}+x^{2}-6 x+5\right)>0$, for all $a \in[-1,2]$ if $f(-1)>0$ or $f(2)>0$. The latter condition is equivalent to $(x+2)(x-1)x>0$. Result: $x \in(-\infty,-2) \cup(0,1) \cup(1, \infty)$.