Solve {l}{{(2x^{4}+4)}{(x^{4}-8)}=2x^8-12x^4-32}{y={(2x^4+4)}{(x^4-8)}}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { {(2 x ^ {4} + 4)} {(x ^ {4} - 8)} = 2 x ^ {8} - 12 x ^ {4} - 32 }\\ { y = {(2 x ^ {4} + 4)} {(x ^ {4} - 8)} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$x\in (-\infty,-\sqrt[4]{3}]\cup [\sqrt[4]{3},\infty)\text{, }y=z\text{, }z=2(x^{4}-8)(x^{4}+2)\text{; }x=-\sqrt[4]{3}\approx -1.316074013\text{, }y=-50\text{, }z=-50\text{; }x=\sqrt[4]{3}\approx 1.316074013\text{, }y=-50\text{, }z=-50\text{; }x=0\text{, }y=-32\text{, }z=-32\text{; }x=-\frac{2^{\frac{3}{4}}\sqrt[4]{-\sqrt{2(z+50)}+6}}{2}\text{, }y=z\text{, }z\in [-50,-32]\text{; }x=\frac{2^{\frac{3}{4}}\sqrt[4]{-\sqrt{2(z+50)}+6}}{2}\text{, }y=z\text{, }z\in [-50,-32]$$