Not sure where I need to go to complete:
$$\frac 49 \times \sqrt{\frac{27}8}.$$
I can get to $(2/3)^{5/2}$ but also not sure if this is correct.
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You have
$$\frac 49\sqrt{\frac{27}8}=\frac 49\times\frac{3\sqrt{3}}{2\sqrt{2}}=\frac 23 \times \sqrt{\frac 32}=\sqrt{\frac 23}=\left(\frac 23\right)^{1/2}.$$
E. Joseph
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$$\frac{4}{9} \cdot \sqrt{\frac{27}{8}}=\frac{4}{9}\cdot \frac{3}{2}\sqrt{\frac{3}{2}}=\frac{2}{3}\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{3}$$
But $\frac{2}{3}^{5/2}=\frac{4}{9}\sqrt{\frac{2}{3}}$ , which is obviously another number because $\sqrt{\frac{27}{8}}>1>\sqrt{\frac{2}{3}}$
Peter
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...and $;\frac{\sqrt6}3=\sqrt\frac23;$ . Which one is "better" is, I believe, subjective. – DonAntonio Sep 30 '16 at 21:40
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Normally, the task is to rationalize the denominator , so I chose this way. – Peter Sep 30 '16 at 21:41
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Perhaps. I honestly don't know. I guess that's a high school thing. – DonAntonio Sep 30 '16 at 21:43
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It is actually a matter of taste. Your variant has the advantage that it is easier to see that $\frac{4}{9}\sqrt{\frac{2}{3}}$ cannot be corrrect. – Peter Sep 30 '16 at 21:43
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1Yes, that was my point. +1 – DonAntonio Sep 30 '16 at 21:44