The following equation needs to be solved:

$$
x+\sqrt[8]{x^{5}}-12 \sqrt[4]{x}=0
$$
$x=y^{8}$ substitution

$$
y^{8}+y^{5}-12 y^{2}=0
$$

Factoring out $y^{2}$

$$
y^{2}\left(y^{6}+y^{3}-12\right)=0
$$

So either $y=0$, or $y^{6}+y^{3}-12=0$.

The latter is a quadratic equation in terms of $y^{3}$, whose solutions are

$$
y_{1}^{3}=3 \quad \text { and } \quad y_{2}^{3}=-4, \quad y_{1}=\sqrt[3]{3}, \quad y_{2}=-\sqrt[3]{4}
$$

and thus

$$
x_{1}=3^{\frac{8}{3}}, \quad x_{2}=4^{\frac{8}{3}} \quad \text { and } \quad x_{3}=0
$$

Of these, only $3^{\frac{8}{3}}=9 \sqrt[3]{9}$ and 0 satisfy the original equation.