AustriaMO2011

The first equation can be transformed as follows: $$\begin{array}{rl}& 2^{\sqrt[3]{x^2}}\cdot 4^{\sqrt[3]{y^2}}\cdot 16^{\sqrt[3]{z^2}}=128\\ \iff & 2^{\sqrt[3]{x^2}+2\cdot \sqrt[3]{y^2}+4\cdot \sqrt[3]{z^2}}=2^7\\ \iff & \sqrt[3]{x^2}+2\sqrt[3]{y^2}+4\sqrt[3]{z^2}=7\end{array}$$

and the second as $$\begin{array}{rl}(x{y}^2+z^4{)}^2& =4+(x{y}^2-z^4{)}^2\\ \iff x^2{y}^4+2x{y}^2{z}^4+z^8& =4+(x^2{y}^4-2x{y}^2{z}^4+z^8)\\ \iff 4x{y}^2{z}^4=4& \\ \iff x{y}^2{z}^4=1.& \end{array}$$ Because of the second equation, we note that $x$ must be positive, and we therefore have $x=|x|$. Since the values of two different means $m_0$ and $m_{2/3}$ of the same variables are equal, each variable must be equal to the mean value 1. We therefore have $x=|y|=|z|=1$, and the set of solutions of the given system of equations is therefore $(1,-1,-1)$, $(1,-1,1)$, $(1,1,-1)$, $(1,1,1)$. qed