Question

(-x^3/27y^-4)^2/3

step by step?

Answer 1

\[(-(x{}^{\wedge}3/(27/y{}^{\wedge}4))){}^{\wedge}2/3 \]\[\left(-\left(x^3/\left(27\left/y^4\right.\right)\right)\right){}^{\wedge}\left(\frac{2}{3}\right) \]\[\left(-\left(\frac{x^3 y^4}{27}\right)\right){}^{\wedge}\left(\frac{2}{3}\right) \]\[\sqrt[3]{\left(-\left(\frac{x^3 y^4}{27}\right)\right)^2} \]\[\sqrt[3]{\left(\frac{x^6 y^8}{729}\right)} \]\[\sqrt[3]{\frac{1}{729}}\sqrt[3]{x^6 y^8} \]\[\frac{1}{9}\sqrt[3]{x^6 y^8} \text{ or } \frac{1}{9}\left(x^6 y^8\right)^{1/3} \]

Answer 2

\[(x^6)^{1/3}=x^2\]
So
\[\frac{1}{9}(x^6y^8)^{1/3}=\frac{x^2}{9}*y^{\frac{8}{3}}\]

Answer 3

the answer my teacher gave me is x^2y^4/9

How does that work?

Answer 4

If the original question had \[y^{-6}\] not \[y^{-4}\] then the answer would be the one your teacher gave you. All of the above working would be the same.

Answer 5

sorry y^-6

Answer 6

WolframAlpha.com does not raise the expression to the 2/3 power as I did. Wolfram raised the expression to the 2nd power and then divided the expression by 3. Wolfram adheres to the normal precedence rules.

http://www.wolframalpha.com/input/?i=simplify+(-x%5E3%2F27y%5E-4)%5E2%2F3&t=sftb01

Answer 7

\[\frac{\left(-\left(x^3/\left(27\left/y^4\right.\right)\right)\right){}^{\wedge}2}{3}=\frac{x^6 y^8}{2187} \]

Answer 8

sorry y^-6

Answer 9

i got it.

Answer 10

\[\frac{\left(-\left(x^3/\left(27\left/y^4\right.\right)\right)\right){}^{\wedge}2}{3} \]\[\frac{\left(-\left(x^3/\left(\frac{27}{y^4}\right)\right)\right){}^{\wedge}2}{3} \]\[\frac{\left(-\left(\frac{x^3 y^4}{27}\right)\right){}^{\wedge}2}{3} \]\[\frac{\frac{x^6 y^8}{729}}{3} \]\[\frac{x^6 y^8}{2187} \]I believe it's correct now.

Answer 11

no its not. just stop.