$\displaystyle \sqrt{\frac{4a^3}{27b^3}}$

$\displaystyle =\sqrt{\frac{2\cdot 2\cdot a\cdot a\cdot a}{9\cdot 3\cdot b\cdot b\cdot b}}$ and I can take the square root of 9.

$\displaystyle =6ab\sqrt{\frac{a}{b}}$ is this correct?

The book I'm working from doesn't have the answers so I'm cross checking my work with Wolfram but Wolfram has a habbit of providing the same answer but in a totally different format. If I'm wrong please explain why thank you

This is the solution from Wolfram $\displaystyle \frac{2\sqrt{\frac{a^3}{b^3}}}{3\sqrt{3}}$

**EDIT** as I relook over the problem, maybe the answer might be $\displaystyle \frac{2a}{3b}\sqrt{\frac{a}{3b}}$ ?

2. ## Re: Simplifiy this radical

Originally Posted by uperkurk
$\displaystyle \sqrt{\frac{4a^3}{27b^3}}$

$\displaystyle =\sqrt{\frac{2\cdot 2\cdot a\cdot a\cdot a}{9\cdot 3\cdot b\cdot b\cdot b}}$ and I can take the square root of 9.
You are correct up to here then you get the fractions confused

Collect pairs of the same terms together so you can get the square root easily
$\displaystyle =\sqrt{\frac{2^2\cdot a^2 \cdot a}{3^2\cdot 3\cdot b^2\cdot b}}$

Then split the square root into those terms which are squared and those which aren't
$\displaystyle =\sqrt{\frac{2^2\cdot a^2}{3^2\cdot b^2}}\sqrt{\frac{a}{b\cdot 3}}$

$\displaystyle =\frac{2\cdot a}{3\cdot b}\sqrt{\frac{a}{b\cdot 3}}$

3. ## Re: Simplifiy this radical

Originally Posted by Shakarri

$\displaystyle =\frac{2\cdot a}{3\cdot b}\sqrt{\frac{a}{b\cdot 3}}$
Thanks for your help, but is this not the exact same thing I have?

4. ## Re: Simplifiy this radical

Ah yes it is, as I was posting you edited your post saying that you had looked over the problem again. Wolfram Alpha just expresses the answer differently.

5. ## Re: Simplifiy this radical

Originally Posted by Shakarri
Ah yes it is, as I was posting you edited your post saying that you had looked over the problem again. Wolfram Alpha just expresses the answer differently.
ok so I'm looking at this one, hopefully you can point me in the right direction.

$\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}$

I'll say I need to do something like this $\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}} = \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}}$ so now I have the denominators all the same, now what? I could take the cube root of most things?

$\displaystyle \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} = \frac{\sqrt[4]{6\cdot 9\cdot a^8\cdot b}}{\sqrt{3bc}}$ I'm lost here...

**EDIT** nah that surely isn't right, the 4th root is making it confusing

6. ## Re: Simplifiy this radical

Originally Posted by uperkurk
ok so I'm looking at this one, hopefully you can point me in the right direction.

$\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}$

I'll say I need to do something like this $\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}} = \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}}$ so now I have the denominators all the same, now what? I could take the cube root of most things?

$\displaystyle \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} = \frac{\sqrt[4]{6\cdot 9\cdot a^8\cdot b}}{\sqrt{3bc}}$ I'm lost here...

**EDIT** nah that surely isn't right, the 4th root is making it confusing
It shouldn't as it is true that:
$\displaystyle \sqrt[R]{A} = {A^{\frac{1}{R}}}$

7. ## Re: Simplifiy this radical

I'll say I need to do something like this $\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}}$
There should be no reason to multiply by $\displaystyle \sqrt[4]{\frac{27b}{27b}}$ if you are going to simplify it immediately. You could simplify it before bringing them into 1 fraction
$\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}}=\sqrt[4]{\frac{2a^8}{b^2c^3}}\times 1$

Learn the rules of indices for taking roots. When you have $\displaystyle \sqrt[4]{a^3}$ that is equal to $\displaystyle (a^3)^\frac{1}{4}$ which in turn is equal to $\displaystyle a^\frac{3}{4}$

8. ## Re: Simplifiy this radical

Originally Posted by uperkurk
ok so I'm looking at this one, hopefully you can point me in the right direction.

$\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}$

I'll say I need to do something like this $\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}} = \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}}$ so now I have the denominators all the same, now what? I could take the cube root of most things?

$\displaystyle \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} = \frac{\sqrt[4]{6\cdot 9\cdot a^8\cdot b}}{\sqrt{3bc}}$ I'm lost here...

**EDIT** nah that surely isn't right, the 4th root is making it confusing
You want to 'rationalize the denominator' , that means the FACTORS of the denominator should become perfect 4th powers.

$\displaystyle \sqrt[4]{ \frac{2a^8}{b^2c^3} \cdot \frac{b^2c}{b^2c}} \ = \ \frac{ \sqrt[4]{2a^8b^2c}}{ \sqrt[4]{b^4c^4}} \ = \ \frac{a^2 \sqrt[4]{2b^2c}}{bc}$