2 problems, similar to each other

[silverlining326]

^3√-5x^4
^3√40xy^6

and

^5√64x^6

thanks

 

[pka]

Is your problem \(\displaystyle \L
\frac{{\sqrt[3]{{ - 5x^4 }}}}{{\sqrt[3]{{40xy^6 }}}}\quad \mbox{or}\quad \frac{{\sqrt[3]{{ - 5}}x^4 }}{{\sqrt[3]{{40}}\left( {xy^6 } \right)}}\)?

Please learn to use grouping symbols!

 

[silverlining326]

Is your problem \(\displaystyle \L
\frac{{\sqrt[3]{{ - 5x^4 }}}}{{\sqrt[3]{{40xy^6 }}}}\quad \mbox\)?

Please learn to use grouping symbols!

yep, thats it...sorry =]

 

[soroban]

Hello, silverlining326!

I'll take a guess at what you meant . . .

\(\displaystyle \L\frac{\sqrt[3]{-5x^4}}{\sqrt[3]{40xy^6}}\)

We have: \(\displaystyle \L\:\sqrt[3]{\frac{-5x^4}{40xy^6}} \;=\;\sqrt[3]{\frac{-x^3}{8y^6}} \;=\;\frac{\sqrt[3]{-x^3}}{\sqrt[3]{8y^6}} \;=\;\frac{-x}{2y^2}\)

\(\displaystyle \L\sqrt[5]{64x^6}\)

We have: \(\displaystyle \L\:\sqrt[5]{32\,\cdot\,2\,\cdot\,x^5\,\cdot\,x} \:=\:\sqrt[5]{2^5\cdot x^5\,\cdot\,2x} \;=\;\sqrt[5]{2^5\cdot x^5}\,\cdot\,\sqrt[5]{2x} \;= \;2x\,\sqrt[5]{2x}\)