Urgent! Please help me simplify these expressions. I have no idea. Please show how each term is simplified. thank you. :)

http://img183.imageshack.us/img183/2280/math1.jpg

http://img197.imageshack.us/img197/2851/math2g.jpg

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- Sep 15th 2009, 07:35 AMreiwardRadical Expressions
Urgent! Please help me simplify these expressions. I have no idea. Please show how each term is simplified. thank you. :)

http://img183.imageshack.us/img183/2280/math1.jpg

http://img197.imageshack.us/img197/2851/math2g.jpg - Sep 15th 2009, 12:22 PMmasters
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.

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

$\displaystyle \frac{z^3 \cdot 2^{\frac{1}{5}}x^{\frac{3}{5}}y^{\frac{2}{5}}}{2^{ \frac{1}{2}}x^{\frac{1}{2}}yz^2}\cdot \frac{2^{\frac{1}{2}}x^{\frac{1}{2}}y^{\frac{3}{5} }}{2\cdot 2^{\frac{1}{5}}x\cdot x^{\frac{3}{5}}}=\frac{zy}{2xy}=\frac{z}{2x}$ - Sep 15th 2009, 02:20 PMreiwardthanks!
thank you! maybe thats the final answer for number 1. oh well, thank you very much :)

- Sep 17th 2009, 08:42 AMreiward
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

$

so the answer is.

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

$