# Math Help - [SOLVED] division of an algebric expression - no 2 (and last)

1. ## [SOLVED] division of an algebric expression - no 2 (and last)

Here I go one more that I'm really not sure of, what I really hate is that we have all that practice homework but none of the exercise where as complexe as those in the actual homework, so I basically have no reference...

$

\left( \frac {x^{-1}(x-y)^3 \sqrt[4]{y^3}}{x^{-4}y \sqrt {(x-y)}} \right)^{-2}

$

$\left( x^{-1-(-4)}(x-y)^{\frac {3}{1}-\frac {1}{2}}(y^{1-3})^\frac {1}{4} \right)^{-2}
$

$\left( x^3(x-y)^{\frac {6}{2}-\frac {1}{2}}(y^2)^\frac {1}{4} \right)^{-2}
$

$\left( x^3(x-y)^{\frac {5}{2}}(y^{2^(\frac {1}{4})}) \right)^{-2}
$

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

$x^{-6}(x-y)^{-5}y^{-1}$

Can I simplify this even further? The answer cannot contain fractions or radicals.

Thank you so much, for all the help, I'm eternaly greatful!

2. You probably want to express it without negative powers.

It should simplify to $\frac{y^{\frac{1}{2}}}{x^{6}(x-y)^{5}}$

3. Hello, Neenoon!

Your work is correct . . . This is the way I organize the parts.

$\left[\frac {x^{-1}(x-y)^3 \sqrt[4]{y^3}}{x^{-4}y \sqrt {(x-y)}} \right]^{-2}$

We have: . $\left[\frac{x^{\text{-}1}}{x^{-4}}\cdot \frac{\sqrt[4]{y^3}}{y}\cdot\frac{(x-y)^3}{\sqrt{x-y}}\right]^{-2}$ . $= \;\left(\frac{x^{-1}}{x^{-4}}\cdot\frac{y^{\frac{3}{4}}}{y^1}\cdot\frac{(x-y)^3}{(x-y)^{\frac{1}{2}}} \right)^{-2}$

. . $= \;\left[x^{-1-(-4)} \cdot y^{\frac{3}{4}-1} \cdot(x-y)^{3-\frac{1}{2}}\right]^{-2} \;=\;\left[x^3\cdot y^{-\frac{1}{4}}\cdot(x-y)^{\frac{5}{2}}\right]^{-2}$

. . $= \;\left[x^3\right]^{\text{-}2}\cdot\left[y^{-\frac{1}{4}}\right]^{-2}\cdot\left[(x-y)^{\frac{5}{2}}\right]^{-2} \;=\;x^{-6}\cdot y^{\frac{1}{2}}\cdot(x-y)^{-5}$

. . $=\;\frac{\sqrt{y}}{x^6\,(x-y)^5}$

4. Isn't this step wrong because the final part should be sqrt y? Also he said his answer could not contain radicals or fractions, so how would he get rid of the radical sqrt(y)?

5. Originally Posted by 11rdc11
Isn't this step wrong because the final part should be sqrt y? Also he said his answer could not contain radicals or fractions, so how would he get rid of the radical sqrt(y)?
Just stop at this step then:

$\;\left[x^3\right]^{\text{-}2}\cdot\left[y^{-\frac{1}{4}}\right]^{-2}\cdot\left[(x-y)^{\frac{5}{2}}\right]^{-2} \;=\;x^{-6}\cdot y^{\frac{1}{2}}\cdot(x-y)^{-5}
$

6. Originally Posted by Neenoon
Here I go one more that I'm really not sure of, what I really hate is that we have all that practice homework but none of the exercise where as complexe as those in the actual homework, so I basically have no reference...

$

\left( \frac {x^{-1}(x-y)^3 \sqrt[4]{y^3}}{x^{-4}y \sqrt {(x-y)}} \right)^{-2}

$

$\left( x^{-1-(-4)}(x-y)^{\frac {3}{1}-\frac {1}{2}}{\color{red}(y^{\frac{3}{4}-1})} \right)^{-2}
$

$\left( x^3(x-y)^{\frac {6}{2}-\frac {1}{2}}{\color{red}(y^{-\frac{1}{4}})} \right)^{-2}
$

$\left( x^3(x-y)^{\frac {5}{2}} {\color{red}(y^{-\frac{1}{4}})} \right)^{-2}
$

$\left( x^3(x-y)^{\frac {5}{2}}{\color{red}y^{-\frac {1}{4}} }\right)^{-2}
$

$x^{-6}(x-y)^{-5}{\color{red}y^{\frac{1}{2}}}$

Can I simplify this even further? The answer cannot contain fractions or radicals.

Thank you so much, for all the help, I'm eternaly greatful!
Soroban, your work is impeccable. I have made corrections in the original poster's work. There was a slight problem with the y variable.

7. You guys are hot!

What would I do without you!