Question

Simplify and write in radical form:
((x^3 * y^-2) / xy)^-(1/5)

Answer 1

The answer is $\sqrt[5]{\frac{x^{2}}{y^{3}}}$

$( \frac{ x^{3}* y^{-2} }{xy})^{ \frac{1}{5} } \\ \\ \frac{ x^{a} }{ x^{b} } = x^{a-b} \\ \\ ( \frac{ x^{3}* y^{-2} }{xy})^{ \frac{1}{5} } = ( x^{3-1}y^{-2-1} ) ^{ \frac{1}{5} } = ( x^{2}y^{-3} ) ^{ \frac{1}{5} } \\ \\ x^{-a} = \frac{1}{ x^{a} } \\ \\ ( x^{2}y^{-3} ) ^{2}= (\frac{x^{2}}{y^{3}} )^{ \frac{1}{5} } \\ \\ x^{ \frac{a}{b} } = \sqrt[b]{ x^{a} } \\ \\ (\frac{x^{2}}{y^{3}} )^{ \frac{1}{5} } = \sqrt[5]{\frac{x^{2}}{y^{3}}}$

Answer 2

Answer:

$\sqrt[5]{(\frac{y^3}{x^2})}$

Step-by-step explanation:

$(\frac{x^3\times y^{-2}}{xy})^{\frac{-1}{5}}$

Now we will use the quotient rule of exponent which is given by

$\frac{x^a}{x^b}=a^{a-b}$

Using this property, we get

$({x^{3-1}\times y^{-2-1}})^{\frac{-1}{5}}$

On simplifying, we get

$({x^{2}\times y^{-3}})^{\frac{-1}{5}}$

Now, use the property, $x^{-a}=\frac{1}{x^a}$

$(\frac{x^2}{y^3})^{\frac{-1}{5}}$

Now, we can use the rule $x^{1/a}=\sqrt[a]{x}$

$(\frac{y^3}{x^2})^{\frac{1}{5}}\\\\=\sqrt[5]{(\frac{y^3}{x^2})}$