An algebra problem by ismet cosic

Considering the third equation, $x,y,z > 0$ . $(0,0,0)$ isn't a solution because $0^{0}$ is not defined.

$y^{z} =x^{\frac{2}{3}}$

$y^{z\times z}=x^{\frac{2}{3} \times z} = (x^{z})^{\frac{2}{3}}= (y^{\frac{8}{3}})^{\frac{2}{3}} = y^{\frac{16}{9}}$

$y^{z^{2}} = y^{\frac{16}{9}}$ This means that either $y=1$ or $z^{2}=\frac{16}{9}$ . For $y=1$ we get the first given set of solutions $(1,1,2)$ .

For $z^{2}=\frac{16}{9}$ we get $z=\frac{4}{3}$ , meaning $c=\frac{4}{3}$ .

From the first equation, we get $x^{\frac{4}{3}} = y^{\frac{8}{3}}$ , or $x = y^{2}$ .

From this point, we could solve for x and y, but that's unnecessary because we know that $a = b^{2}$ .

$\frac{\sqrt{a}}{b} \times c=\frac{\sqrt{b^{2}}}{b} \times \frac{4}{3}= \frac{b}{b} \times \frac{4}{3}= 1 \times \frac{4}{3}= \frac{4}{3} =\boxed{1.333}$