# Math Help - Proving Horrible exponent

1. ## Proving Horrible exponent

Hi All,

Prove

$\frac{a}{ (\sqrt[3]{ax}}^2 = \frac{\sqrt[3]{ax}{x}$

$\frac{a}{ (ax)^{1/3} )^2 = \frac{a}{(ax)^{2/3} }$

$\frac{a}{ (ax)^{1/3} \times (ax)^{1/3}}$

Had Latex trouble (again); question is in attachment below. Thanks

2. ## Re: Proving Horrible exponent

Originally Posted by BobBali
Prove:

$\frac{a}{\sqrt[3]{ax}^2} = \frac{\sqrt[3]{ax}}{x}$

$\frac{a}{ \left((ax)^{1/3} \right)^2} = \frac{a}{ (ax)^{2/3} }}$

$\frac{a}{ (ax)^{1/3} \times (ax)^{1/3}}$
I fixed the LaTeX only! (Not sure if this was meant ....?)

3. ## Re: Proving Horrible exponent

Yes it is! How did you do it? My latex working in the second step (the one u fixed) is below, can you tell me where i went wrong?? Thanks.

$\frac{a}{ ( (\sqrt[3]{ax}) )^2 = \frac{a}{ ( (ax)^{1/3} )^2$

4. ## Re: Proving Horrible exponent

Originally Posted by BobBali
Yes it is! How did you do it? My latex working in the second step (the one u fixed) is below, can you tell me where i went wrong?? Thanks.

$\frac{a}{ ( (\sqrt[3]{ax}) )^2 = \frac{a}{ ( (ax)^{1/3} )^2$
\frac{a}{ \left(\sqrt[3]{ax} \right)^2 } = \frac{a}{ \left( (ax)^{1/3} \right)^2 }

yields

$\frac{a}{ \left(\sqrt[3]{ax} \right)^2 } = \frac{a}{ \left( (ax)^{1/3} \right)^2 }$

I've marked the changes and additions in red.

5. ## Re: Proving Horrible exponent

Ok, so:

$\frac{a}{ \left (ax)^{2/3} \right }$ =
$\frac{a}{ \left(ax)^{1/3} \times (ax)^{1/3} \right)$

$a \times a^{-1/3} \times x^{-1/3} \times a^{-1/3} \times x^{-1/3} =$
$a^{1/3} \times x^{-2/3}$ =
$\frac{\sqrt[3]{a}{x}^{2/3}$

6. ## Re: Proving Horrible exponent

Originally Posted by BobBali
Ok, so:

$\frac{a}{ \left (ax)^{2/3} \right }$ =
$\frac{a}{ \left(ax)^{1/3} \times (ax)^{1/3} \right)}$

$a \times a^{-1/3} \times x^{-1/3} \times a^{-1/3} \times x^{-1/3} =$
$a^{1/3} \times x^{-2/3}$ =
$\frac{\sqrt[3]{a}}{x^{2/3}}$

Please use the \frac-command in such a way:

\frac{numerator}{denominator} yields $\frac{numerator}{denominator}$

7. ## Re: Proving Horrible exponent

Hello, BobBali!

$\text{Prove: }\:\frac{a}{(\sqrt[3]{ax})^2} \:=\:\frac{\sqrt[3]{ax}}{x}$

On the left side, we have: . $\frac{a}{(ax)^{\frac{2}{3}}}$

Multiply by $\tfrac{(ax)^{\frac{1}{3}}}{(ax)^{\frac{1}{3}}}: \;\;\frac{a}{(ax)^{\frac{2}{3}}} \cdot\frac{(ax)^{\frac{1}{3}}}{(ax)^{\frac{1}{3}}} \;=\;\frac{a(ax)^{\frac{1}{3}}}{ax} \;=\;\frac{(ax)^{\frac{1}{3}}}{x} \;=\;\frac{\sqrt[3]{ax}}{x}$