If ${\mathrm{x}}^{2/3}-7{\mathrm{x}}^{1/3}+10=0, \mathrm{then} \mathrm{x}=$

Given that $x^{2/3}-7x^{1/3}+10=0.$

Given equation can be written as

${\left({\mathrm{x}}^{1/3}\right)}^2-7\left({\mathrm{x}}^{1/3}\right)+10=0$

Let $\mathrm{a}={\mathrm{x}}^{1/3}$, then it reduces to the equation

$$\begin{gathered} {\mathrm{a}}^2-7\mathrm{a}+10=0 \\ \Rightarrow (\mathrm{a}-5)(\mathrm{a}-2)=0 \\ \Rightarrow \mathrm{a}=5,2 \end{gathered}$$

Putting these values, we have

${\mathrm{a}}^3=\mathrm{x}\Rightarrow \mathrm{x}=125 \mathrm{and} 8$