# How do you simplify (x^(1/pi)/y^(2/pi))^pi?

Jul 21, 2018

${\left({x}^{\frac{1}{\pi}} / {y}^{\frac{2}{\pi}}\right)}^{\pi} = \frac{x}{y} ^ 2$

#### Explanation:

Note that if $a > 0$ and $b , c$ are any real numbers then:

${\left({a}^{b}\right)}^{c} = {a}^{b c}$

So (assuming $x , y > 0$) we find:

${\left({x}^{\frac{1}{\pi}} / {y}^{\frac{2}{\pi}}\right)}^{\pi} = {\left({x}^{\frac{1}{\pi}}\right)}^{\pi} / {\left({y}^{\frac{2}{\pi}}\right)}^{\pi} = \frac{{x}^{\frac{1}{\pi} \cdot \pi}}{{y}^{\frac{2}{\pi} \cdot \pi}} = \frac{x}{y} ^ 2$

Jul 21, 2018

$\frac{x}{y} ^ 2$

#### Explanation:

${\left({x}^{\frac{1}{\pi}} / {y}^{\frac{2}{\pi}}\right)}^{\pi} = {x}^{\frac{\pi}{\pi}} / {y}^{\frac{2 \pi}{\pi}} = \frac{x}{y} ^ 2$

$\setminus \frac{x}{{y}^{2}}$

#### Explanation:

${\left(\setminus \frac{{x}^{\frac{1}{\setminus} \pi}}{{y}^{\frac{2}{\setminus} \pi}}\right)}^{\setminus} \pi$

=\frac{(x^{1/\pi})^\pi}{(y^{2/\pi})^\pi

=\frac{x^{\pi\cdot 1/\pi}}{y^{\pi\cdot 2/\pi}

$= \setminus \frac{{x}^{1}}{{y}^{2}}$

$= \setminus \frac{x}{{y}^{2}}$