# Square root /w fraction and exponent: how to simplilfy correctly

#### Quasimoto

##### New member
Dear forum users,

I've been trying to figure this out.

Could someone explain to me how this is the answer?

#### tkhunny

##### Moderator
Staff member
$$\displaystyle \left(\dfrac{2\sqrt{3}}{3\sqrt{2}}\right)^{3} = \left(\dfrac{2}{3}\right)^{3}\cdot\left(\dfrac{\sqrt{3}}{\sqrt{2}}\right)^{3} = \left(\dfrac{2}{3}\right)^{3}\cdot\left(\dfrac{\sqrt{3}}{\sqrt{2}}\right)^{2}\cdot\left(\dfrac{\sqrt{3}}{\sqrt{2}}\right)$$

Follow?

#### topsquark

##### Full Member
1st expression = ~.5443
2nd expression = ~2.0528
As noted by the expansion that tkhunny is using the "solution" given in the OP is incorrect. The actual answer has a $$\displaystyle \sqrt{2}$$ in the denominator. The fraction in front of the radicals is wrong as well.

Is this the solution to a different problem?

-Dan

#### Quasimoto

##### New member
I
$$\displaystyle \left(\dfrac{2\sqrt{3}}{3\sqrt{2}}\right)^{3} = \left(\dfrac{2}{3}\right)^{3}\cdot\left(\dfrac{\sqrt{3}}{\sqrt{2}}\right)^{3} = \left(\dfrac{2}{3}\right)^{3}\cdot\left(\dfrac{\sqrt{3}}{\sqrt{2}}\right)^{2}\cdot\left(\dfrac{\sqrt{3}}{\sqrt{2}}\right)$$

Follow?
I am following it yes.

#### Quasimoto

##### New member
As noted by the expansion that tkhunny is using the "solution" given in the OP is incorrect. The actual answer has a $$\displaystyle \sqrt{2}$$ in the denominator. The fraction in front of the radicals is wrong as well.

Is this the solution to a different problem?

-Dan
I looked for the answer again and yes the solution is to a different problem.
But my original answer is still wrong.
So I will be trying to figure it out later today.

Thank you

#### pka

##### Elite Member
Dear forum users, I've been trying to figure this out.
You posted two images:
$$\displaystyle \boxed{\left(\dfrac{2\sqrt3}{3\sqrt2}\right)^3}$$ & $$\displaystyle \boxed{\frac{32}{27}\sqrt3}$$
The second image is not the reduction of the first.
You tell us at what step you fail to understand.
1) $$\displaystyle \frac{2\sqrt3}{3\sqrt2}=\frac{2\sqrt3\cdot\sqrt2}{3\sqrt2\cdot\sqrt2}$$
2) $$\displaystyle \frac{2\sqrt3}{3\sqrt2}=\frac{2\sqrt6}{3\cdot 2}$$
3) $$\displaystyle \frac{2\sqrt3}{3\sqrt2}=\frac{\sqrt6}{3}$$

4) $$\displaystyle \left(\frac{\sqrt6}{3}\right)^3=\frac{6\sqrt6}{27}$$
5) $$\displaystyle \boxed{\left(\frac{\sqrt6}{3}\right)^3=\frac{2\sqrt6}{9}}$$ CLICK HERE

#### Quasimoto

##### New member
You posted two images:
$$\displaystyle \boxed{\left(\dfrac{2\sqrt3}{3\sqrt2}\right)^3}$$ & $$\displaystyle \boxed{\frac{32}{27}\sqrt3}$$
The second image is not the reduction of the first.
You tell us at what step you fail to understand.
1) $$\displaystyle \frac{2\sqrt3}{3\sqrt2}=\frac{2\sqrt3\cdot\sqrt2}{3\sqrt2\cdot\sqrt2}$$
2) $$\displaystyle \frac{2\sqrt3}{3\sqrt2}=\frac{2\sqrt6}{3\cdot 2}$$
3) $$\displaystyle \frac{2\sqrt3}{3\sqrt2}=\frac{\sqrt6}{3}$$

4) $$\displaystyle \left(\frac{\sqrt6}{3}\right)^3=\frac{6\sqrt6}{27}$$
5) $$\displaystyle \boxed{\left(\frac{\sqrt6}{3}\right)^3=\frac{2\sqrt6}{9}}$$ CLICK HERE
The second image is indeed the wrong solution to the problem.
Thank you for explaining step by step.

I failed at step 4, I used the exponent at the beginning instead of solving everything what is inside of the parentheses.

Thank you very much I appreciate the time and effort.