1. ## Square expresion

It seems that the part of my last post that stated my tight sleep as true was indeed a lie. I now have another problem witch will haunt my dreems into infinity, that is iff someone doesn't help me a little!

The problem is:
$\sqrt[3]{81}-\sqrt[6]{9}+\sqrt{12}-\sqrt[6]{27}-\sqrt[3]{3\sqrt3}$
Simplify this as far as possible.

As for my attempt of a solution:
$\sqrt[3]{81}=\sqrt[3]{9*9}=\sqrt[3]{3*3*3*3}=3*\sqrt[3]{3}$

$\sqrt[6]{9}=((3*3)^{1/2})^{1/3}=\sqrt[3]{3}$

$\sqrt{12}=\sqrt{2*2*3}=2*\sqrt{3}$

$\sqrt[6]{27}=\sqrt[6]{3*3*3}=((3*3*3)^{1/3})^{1/2}=\sqrt{3}$

$\sqrt[3]{3*\sqrt{3}}=(3*\sqrt{3})^{1/3}=(3*3^{1/3})^{1/3}=3^{4/9}$

Thus the original expression equals:
$3*\sqrt[3]{3}-\sqrt[3]{3}+2\sqrt{3}-\sqrt{3}-3^{4/9} = }$
$2*\sqrt[3]{3}+\sqrt{3}-3^{4/9} \not= 2*\sqrt[3]{3}$

Where $2*\sqrt[3]{3}$ is the answer given by my textbook.
What did I do wrong?

2. Originally Posted by a4swe
$\sqrt[3]{3*\sqrt{3}}=(3*\sqrt{3})^{1/3}=(3*3^{1/3})^{1/3}=3^{4/9}$
This is where you're off, it should be:

$\sqrt[3]{3*\sqrt{3}}=(3*\sqrt{3})^{1/3}=(3*3^{1/2})^{1/3}=3^{3/6}=\sqrt{3}$

Now let's finish: $\sqrt[3]{81}-\sqrt[6]{9}+\sqrt{12}-\sqrt[6]{27}-\sqrt[3]{3\sqrt3}$

simplify: $3*\sqrt[3]{3}-\sqrt[3]{3}+2\sqrt{3}-\sqrt{3}-\sqrt3$

group like terms: $\left(3*\sqrt[3]{3}-\sqrt[3]{3}\right)+\left(2\sqrt{3}-\sqrt{3}-\sqrt3\right)$

simplify: $\left((3-1)*\sqrt[3]{3}\right)+\left(0\right)$

therefore: $\boxed{2*\sqrt[3]{3}}$

3. Thank you very much, that was quick.
I thought it was some minor error just like that witch I just couldn't spot.
And belive me, I have looked for it, for a long time.