2. Originally Posted by reiward

Hi reiward,

This is truly some industrial strength algebra. I got the first one down fairly simple, but couldn't manipulate the last term to very much. Maybe it's stated incorrectly or something.

$\displaystyle \sqrt[6]{32 \sqrt{4n^{12}}} - \frac{n\sqrt{4n^2+16n+16}}{n+2}+\frac{n-1}{\sqrt[3]{n}-1}$

$\displaystyle \sqrt[6]{32 \cdot 2n^6} - \frac{n\sqrt{(2n+4)^2}}{n+2}+\frac{n-1}{\sqrt[3]{n}-1}$

$\displaystyle \sqrt[6]{2^6 \cdot n^6} - \frac{n(2n+4)}{n+2}+\frac{n-1}{\sqrt[3]{n}-1}$

$\displaystyle 2n - \frac{2n(n+2}{n+2}+\frac{n-1}{\sqrt[3]{n}-1}$

$\displaystyle 2n - 2n +\frac{n-1}{\sqrt[3]{n}-1}$

$\displaystyle \frac{n-1}{\sqrt[3]{n}-1}$

I messed around with this last term a while, but could never get it to anything simpler than it already is.

Now, for the second one.

$\displaystyle \frac{\sqrt{z^6}\cdot \sqrt[5]{2x^3y^2}}{\sqrt[4]{4x^2y^4z^8}} \div \frac{\sqrt[5]{64x^8y^{-3}}}{\sqrt[6]{8x^3}}$

$\displaystyle \frac{z^3 \cdot 2^{\frac{1}{5}}x^{\frac{3}{5}}y^{\frac{2}{5}}}{2^{ \frac{1}{2}}x^{\frac{1}{2}}yz^2}\cdot \frac{2^{\frac{1}{2}}x^{\frac{1}{2}}y^{\frac{3}{5} }}{2\cdot 2^{\frac{1}{5}}x\cdot x^{\frac{3}{5}}}=\frac{zy}{2xy}=\frac{z}{2x}$

3. ## thanks!

thank you! maybe thats the final answer for number 1. oh well, thank you very much

4. hi, we discussed the answers already, just wanna share. for the last term this is what we did.

$\displaystyle \frac{n-1}{\sqrt[3]{n}-1} * \frac{\sqrt[3]{n}^2 + \sqrt[3]{n} + 1}{\sqrt[3]{n}^2 + \sqrt[3]{n} + 1}$

then

$\displaystyle \frac{n-1} {n-1} \sqrt[3]{n}^2 + \sqrt[3]{n} + 1$

$\displaystyle \sqrt[3]{n}^2 + \sqrt[3]{n} + 1$