x^a=y^1/3
y^b=z^1/3
z^c=x^1/3
abc=?

Dear Student,

Please find below the solution to the asked query:

We have : 

$$\begin{gathered} x^{\mathrm{a}} = y^{\frac{1}{3}} , \mathrm{Taking} \mathrm{whole} \mathrm{cube} \mathrm{on} \mathrm{both} \mathrm{hand} \mathrm{side} \mathrm{we} \mathrm{get} : \\ {\left(x^{\mathrm{a}} \right)}^3={\left( y^{\frac{1}{3}}\right)}^3 \\ {x}^{3\mathrm{a}} = y --- ( 1 )\hspace{0.17em} ( \mathrm{We} \mathrm{know} {\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}} = {\mathrm{a}}^{\mathrm{m}\times \mathrm{n}} ) \\ \mathrm{Similarly} \mathrm{for} \mathrm{rest} \mathrm{of} \mathrm{the} \mathrm{equations} \mathrm{we} \mathrm{get} : \\ {y}^{\mathrm{b}} = z^{\frac{1}{3}} \\ {y}^{3b} = z --- ( 2 )\hspace{0.17em} \\ \mathrm{And} \\ {z}^{\mathrm{c}} = x^{\frac{1}{3}} \\ {z}^{3\mathrm{c}} = x --- ( 3 )\hspace{0.17em} \end{gathered}$$

Now we substitute value of ' x ' from equation ( 3 ) in equation ( 1 ) and get :

$$\begin{gathered} {\left(z^{3c}\right)}^{3a} = y \\ \Rightarrow z^{9ac}= y --- ( 4 ) \end{gathered}$$
And we substitute value of ' z ' from equation ( 2 ) in equation ( 4 ) and get :
$$\begin{gathered} {\left(y^{3b}\right)}^{9ac} = y \\ \mathbf{\Rightarrow }{\mathit{y}}^{\mathbf{27}\mathbf{a}\mathbf{b}\mathbf{c}}\mathbf{=}\mathbf{ }\mathit{y} \end{gathered}$$
Now we compare both side and get :

27 abc = 1  ,

abc = $\frac{\mathbf{1}}{\mathbf{27}}$                                                 ( Ans )

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards