Solve the following inequality:

$$
x^{3}+1>x^{2}+x
$$
Let's transform the inequality (1) and factorize it:

$$
x^{3}-x^{2}-x+1=\left(x^{2}-1\right)(x-1)>0
$$

The product is positive if either both factors are positive or both factors are negative.

$(x-1)>0$ if $x>1$, and in this case $x^{2}-1>0$ also holds, so the inequality (1) is true.

If $x=1$, then $x-1=0$, and if $x=-1$, then $x^{2}-1=0$, which is not a solution.

If $-1<x<1$, then $(x-1)<0$ and $x^{2}-1<0$, so the product $\left(x^{2}-1\right)(x-1)>0$ does not hold, hence the inequality (1) does not hold.

If $x<-1$, then $(x-1)<0$ and $x^{2}-1>0$, so the product $\left(x^{2}-1\right)(x-1)<0$, hence the inequality (1) does not hold.

In summary, the solution to the inequality is the set of values $-1<x<1$ and $x>1$.

Czirkos Zoltán (Budapest, ELTE Apáczai Csere J. Gymn., 10th grade)