xy^2=a^3 ,yz^2=b^3 , zx^2=c^3 then what is z^3 equal to?

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} {\mathrm{xy}}^2={\mathrm{a}}^3, {\mathrm{yz}}^2={\mathrm{b}}^3, {\mathrm{zx}}^2={\mathrm{c}}^3 \\ \mathrm{We} \mathrm{need} \mathrm{to} \mathrm{fnd} \mathrm{the} \mathrm{value} \mathrm{of} {\mathrm{z}}^3 \\ \mathrm{From} \mathrm{the} \mathrm{condition} {\mathrm{yz}}^2={\mathrm{b}}^3, \mathrm{bytaking} \mathrm{fourth} \mathrm{power} \mathrm{on} \\ \mathrm{both} \mathrm{sides}, \\ {\left( {\mathrm{yz}}^2\right)}^4={\left({\mathrm{b}}^3\right)}^4 \\ {\mathrm{y}}^4{\mathrm{z}}^8={\mathrm{b}}^{12} ......\left(1\right) \\ \mathrm{Square} \mathrm{both} \mathrm{sides} \mathrm{in} {\mathrm{xy}}^2={\mathrm{a}}^3 \mathrm{to} \mathrm{get}, \\ {\left({\mathrm{xy}}^2\right)}^2={\left({\mathrm{a}}^3\right)}^2 \\ {\mathrm{x}}^2{\mathrm{y}}^4={\mathrm{a}}^6 \\ {\mathrm{y}}^4=\frac{{\mathrm{a}}^6}{{\mathrm{x}}^2} .....\left(2\right) \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ \frac{{\mathrm{a}}^6}{{\mathrm{x}}^2}{\mathrm{z}}^8={\mathrm{b}}^{12} \\ \frac{1}{{\mathrm{x}}^2}{\mathrm{z}}^8=\frac{{\mathrm{b}}^{12}}{{\mathrm{a}}^6} \\ \mathrm{From} 3\mathrm{rd} \mathrm{condition} {\mathrm{zx}}^2={\mathrm{c}}^3, \mathrm{we} \mathrm{get} {\mathrm{x}}^2=\frac{{\mathrm{c}}^3}{\mathrm{z}} \\ \mathrm{This} \mathrm{implies} \mathrm{that}, \\ \frac{\mathrm{z}}{{\mathrm{c}}^3}{\mathrm{z}}^8=\frac{{\mathrm{b}}^{12}}{{\mathrm{a}}^6} \\ {\mathrm{z}}^9=\frac{{\mathrm{b}}^{12} {\mathrm{c}}^3}{{\mathrm{a}}^6} \\ \mathrm{Take} \mathrm{cube} \mathrm{root} \mathrm{on} \mathrm{both} \mathrm{sides} \\ \sqrt[3]{ {\mathrm{z}}^9}=\sqrt[3]{\frac{{\mathrm{b}}^{12} {\mathrm{c}}^3}{{\mathrm{a}}^6}} \\ \sqrt[3]{{\left( {\mathrm{z}}^3\right)}^3}=\sqrt[3]{{\left(\frac{{\mathrm{b}}^4 \mathrm{c}}{{\mathrm{a}}^2}\right)}^3} \\ \mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{z}}^{\mathbf{3}}\mathbf{=}\frac{{\mathbf{b}}^{\mathbf{4}}\mathbf{ }\mathbf{c}}{{\mathbf{a}}^{\mathbf{2}}} \end{gathered}$$

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