Simplify under one radical?

[eric_f]

Hi All,

I am at the end of an optimization problem, and although I arrive at the correct answer of: (V/pi)(V/2pi)^(-2/3) The professor wants the answer in this form: 2*cuberoot(V/2pi). I have no idea where to start the algebraic manipulation...

Could it be as simple as multiplying the bottom and top of the fraction by (V/2pi)^(1/3)? By doing that I am able to cancel factors and group appropriately, but I could be arriving at the answer the wrong way...

 

[soroban]

Hello, eric_f!

I arrive at the correct answer of: \(\displaystyle \dfrac{V}{\pi}\left(\dfrac{V}{2\pi}\right)^{\text{-}\frac{2}{3}}\)

The professor wants the answer in this form: .\(\displaystyle 2\sqrt[3]{\dfrac{V}{2\pi}}\)

I have no idea where to start the algebraic manipulation.

You have: .\(\displaystyle \dfrac{V}{\pi}\left(\dfrac{V}{2\pi}\right)^{\text{-}\frac{2}{3}} \;=\;\dfrac{V}{\pi}\left(\dfrac{2\pi}{V}\right)^{ \frac{2}{3}} \;=\;\dfrac{V}{\pi}\cdot\dfrac{2^{\frac{2}{3}}\pi^{\frac{2}{3}}}{V^{\frac{2}{3}}} \)

. . . . . . . . \(\displaystyle \displaystyle=\;2^{\frac{2}{3}}\cdot \dfrac{V}{V^{\frac{2}{3}}}\cdot\dfrac{\pi^{\frac{2}{3}}}{\pi} \;=\;\dfrac{2^{\frac{2}{3}}V^{\frac{1}{3}}}{\pi^{ \frac{1}{3}}} \;=\;\color{red}{\frac{2^{\frac{1}{3}}}{2^{\frac{1}{3}}}}\cdot \frac{2^{\frac{2}{3}}V^{\frac{1}{3}}}{\pi^{\frac{1}{3}}} \)

. . . . . . . . \(\displaystyle \displaystyle =\;2\cdot\frac{V^{\frac{1}{3}}}{2^{\frac{1}{3}}\pi^{\frac{1}{3}}} \;=\;2\left(\frac{V}{2\pi}\right)^{\frac{1}{3}} \;=\;2\sqrt[3]{\frac{V}{2\pi}}\)

 

[eric_f]

You know that a^(-b/c) = 1 / a^(b/c), right?

I do, and this is what I did:



How wrong is that?

Thank you soroban for the walkthrough!

~Eric

 

[JeffM]

Hi All,

I am at the end of an optimization problem, and although I arrive at the correct answer of: (V/pi)(V/2pi)^(-2/3) The professor wants the answer in this form: 2*cuberoot(V/2pi). I have no idea where to start the algebraic manipulation...

Could it be as simple as multiplying the bottom and top of the fraction by (V/2pi)^(1/3)? By doing that I am able to cancel factors and group appropriately, but I could be arriving at the answer the wrong way...

Here is a way that may be simpler to see than soroban's although both are equally valid

\(\displaystyle \dfrac{V}{\pi} * \left(\dfrac{V}{2 \pi}\right)^{-(2/3)} = \dfrac{V}{\pi} * \left(\dfrac{2 \pi}{V}\right)^{(2/3)} = \dfrac{V}{\pi} * \sqrt[3]{\dfrac{(2 \pi)^2}{V^2}} = \sqrt[3]{\dfrac{V^3 * 4 \pi^2}{\pi^3 * V^2}} = \sqrt[3]{\dfrac{4V}{\pi}} = \sqrt[3]{\dfrac{8V}{2 \pi}} = 2\sqrt[3]{\dfrac{V}{2 \pi}}.\)