Đk $x\le\frac12$
Đặt $a=\sqrt[3]{\frac12+x}\le1;b=\sqrt{\frac12-x}\ge0$$\Rightarrow \begin{cases}a+b=1 \\ a^3+b^2=1 \end{cases}\Leftrightarrow \begin{cases}b=1-a \\ a^3+(1-a)^2=1 \end{cases}$
$\Rightarrow a^3+a^2-2a=0$
$\Leftrightarrow \left[ {\begin{matrix} a=0\\a=1\\a=-2 \end{matrix}} \right. \Leftrightarrow \left[ {\begin{matrix} x=-\frac12(tm)\\ x=\frac12(tm)\\x=-\frac{17}2(tm)\end{matrix}} \right.$