Solve the following equation:

$$
x^{6}-x^{3}-2 x^{2}-1=2\left(x-x^{3}+1\right) \sqrt{x}
$$
Solution. Rearrange the equation and add $x$ to both sides:

$$
\begin{aligned}
x^{6}+2 x^{3.5}+x & =x^{3}+2 x^{2}+2 x^{1.5}+x+2 x^{0.5}+1 \\
\left(x^{3}+x^{0.5}\right)^{2} & =\left(x^{1.5}+x^{0.5}+1\right)^{2}
\end{aligned}
$$

The $\sqrt{x}$ is only defined if $x \geq 0$, in which case it is clear that both the left and right sides are at least 0, and the square function is one-to-one, so we can take the square root:

$$
x^{3}+x^{0.5}=x^{1.5}+x^{0.5}+1, \quad x^{3}-x^{1.5}-1=0
$$

Substitute $y=x^{1.5}-\mathrm{t}: y^{2}-y-1=0$. The roots of the quadratic equation are:

$$
y_{1}=\frac{1-\sqrt{5}}{2}, \quad y_{2}=\frac{1+\sqrt{5}}{2}
$$

The power function $x^{1.5}$ is defined if $x$ is non-negative, in which case the power value, i.e., $y$, is also non-negative. This rules out the first root. From the second root:

$$
x=\left(\frac{1+\sqrt{5}}{2}\right)^{2 / 3} \approx 1.378
$$

This is the only solution to the equation.