For which real numbers $x$ does the following inequality hold:

$$
2\left(x^{3}+1\right)^{2}>\left(x^{2}+1\right)^{3}
$$
The difference between the left and right sides of the inequality can be expressed as follows after expansion:

$$
\begin{aligned}
x^{6}-3 x^{4}+4 x^{3}-3 x^{2}+1 & =(x-1)\left(x^{5}+x^{4}-2 x^{3}+2 x^{2}-x-1\right)= \\
& =(x-1)^{2}\left(x^{4}+2 x^{3}+2 x+1\right)
\end{aligned}
$$

(We used the fact that the sum of the coefficients in both the expanded difference and the 5th-degree polynomial obtained from the first transformation is 0, so $x=1$ is a root of the polynomial, and thus the factor $x-1$ can be factored out). Based on this, we are looking for those $x$ for which (2) is positive. Since the $(x-1)^{2}$ factor is positive everywhere except at $x=1$, where it is 0, the solution to (1) is given by the solution to the inequality (2):

$$
x^{4}+2 x^{3}+2 x+1>0
$$

excluding the point $x=1$.

We further transform (3) by completing the square on the left side, factoring it, and completing the square of its factors:

$$
\begin{aligned}
\left(x^{2}+x+1\right)^{2}-3 x^{2} & =\left[x^{2}+(1+\sqrt{3}) x+1\right]\left[x^{2}+(1-\sqrt{3}) x+1\right]= \\
& =\left[\left(x+\frac{1+\sqrt{3}}{2}\right)^{2}-\frac{\sqrt{3}}{2}\right]\left[\left(x+\frac{1-\sqrt{3}}{2}\right)^{2}+\frac{\sqrt{3}}{2}\right]>0
\end{aligned}
$$

The second factor is positive for all $x$, and the first factor is positive everywhere except in the interval between its roots:

$$
x_{1}=-\frac{1+\sqrt{3}+\sqrt{2 \sqrt{3}}}{2}(\approx-2.297) \quad \text { and } \quad x_{2}=-\frac{1+\sqrt{3}-\sqrt{2 \sqrt{3}}}{2}(\approx-0.435)
$$

(excluding the endpoints as well), since the coefficient of $x^{2}$ in it is positive.

Therefore, the solution to (1) is:

$$
x1
$$

Remark. In solving (3), we can also start from the observation that the coefficients of the terms counted from the front and the back (which form pairs with the sum of exponents being 4) are equal in pairs. Based on this, by introducing the new variable $y=x+\frac{1}{x}, x \neq 0$,

$$
x^{4}+2 x^{3}+2 x+1=x^{2}\left(x^{2}+2 x+\frac{2}{x}+\frac{1}{x^{2}}\right)=x^{2}\left(y^{2}+2 y-2\right)>0
$$

holds if

$$
y^{2}+2 y-2>0, \quad y-1+\sqrt{3}
$$

However, this leads us to (4) indirectly, and we need to separately examine the case $x=0$.