1st expression = ~.5443Dear forum users,
I've been trying to figure this out.
View attachment 12584
Could someone explain to me how this is the answer?
View attachment 12585
Thank you in advance!
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.1st expression = ~.5443
2nd expression = ~2.0528
[math]\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)[/math]
Follow?
I looked for the answer again and yes the solution is to a different problem.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
You posted two images:Dear forum users, I've been trying to figure this out.
The second image is indeed the wrong solution to the problem.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