Not that Complicated

Let $a=\sqrt[3]{z+2}$ and $b=\sqrt[3]{2-z}$ . Then, we know that $a^2+3b^2=4ab$ . $(a-b)(a-3b)=0\implies{a=b}$ or $a=3b$ . By substituting back the values, we obtain $\sqrt[3]{z+2}=\sqrt[3]{2-z}$ or $\sqrt[3]{z+2}=3(\sqrt[3]{2-z})$ . Cubing both sides of the equation results in $z+2=2-z$ or $z+2=27(2-z)$ . $\therefore{z=0}$ or $z=1\frac{6}{7}$ , the required integer solution is $z=0$ .