## Task 25/82

Solve the equation $\sqrt[3]{24}+x+\sqrt[2]{12}-x=6$, where $x$ is a real number!
We set, to eliminate the cube root; $\sqrt[3]{24+x}=a$, thus $24+x=a^{3}, x=a^{3}-24, 12-x=$ $36-a^{3}$. Then the given equation takes the form

$$
a+\sqrt{36-a^{3}}=6 \quad \text { or } \sqrt{36-a^{3}}=6-a
$$

Squaring both sides yields

$$
36-a^{3}=36-12 a+a^{2} \quad ; \quad a\left(a^{2}+a-12\right)=0
$$

with the solutions $a_{1}=0, a_{2,3}=-0.5 \pm 3.5$. This gives $x_{1}=-24, x_{2}=3, x_{3}=-88$. However, only $x_{1}=-24$ and $x_{2}=3$ satisfy the verification. Therefore, $L=\{x \mid x=-24 ; x=3\}$.