Can I Take Its Root First?

$\sqrt [3]{125^2}$ = $\sqrt[3]{{(5^3)}^2}$ . = $\sqrt[3]{5^6}$ .
As we know $\sqrt[n]{a^x} = a^{\frac {x}{n}}$ .
So $\sqrt[3]{5^6} = 5^{\frac {6}{3}}$ .
= $5^2$ = 25

$\sqrt[3]{125^{2}} = 125^{\frac{2}{3}} = 125^{1-\frac{1}{3}} = \frac{125}{\sqrt[3]{125}} = \frac{125}{5} = 25$

$\sqrt[3]{125^2}$ = $\sqrt[3]{(5^3)^2}$ = $\sqrt[3]{5^6}$ = $5^\frac{6}{3}$ = $5^2$ = 25

For some real $m, \, n$ and $z$ , $\sqrt[m]{ z^n }={ \left( \sqrt[m]{z} \right) }^n$ . Therefore: $\sqrt[3]{ 125^2 }={ \left( \sqrt[3]{125} \right) }^2={ \left( \sqrt[3]{125} \right) }^2=5^2= \boxed{\boxed{25}}$

• Try this one as well..

(125x125)^(1/3)→»→»(5^3x5^3)^(1/3)

→→→»5x5=25