11.2 Solve the equation $|x+1|+|x-1|=2 x^{3}$.
11.2 We have $|x+1|=0$ when $x=-1$, and $|x-1|=0$ when $x=1$. We will solve the given equation in the intervals $(-\infty,-1)$, $[-1,1)$, $[1, \infty)$.

1) If $x<0$ for any $x$, then $x=0$; this value does not belong to the interval $(-\infty,-1)$.
2) If $-1 \leq x<1$, then $x+1-x+1=2 x^{3} ; 2 x^{3}=2 ; x^{3}=1$; $x=1 \notin[-1,1)$.
3) If $\quad x \geq 1, \quad$ then $\quad x+1+x-1=2 x^{3} ; \quad x^{3}-x=0$; $x\left(x^{2}-1\right)=0 ; x(x-1)(x+1)=0 ; \quad x_{1}=-1, x_{2}=0, x_{3}=1$. Only the value $x=1$ belongs to the interval $[1, \infty)$. Therefore: $x=1$.

Answer: 1.