Examples

You use the rules for roots, which gives you $\sqrt[3]{a^2}=a^{\frac{2}{3}}$, $\sqrt{a}=a^{\frac{1}{2}}$, and $\sqrt[6]{a^7}=a^{\frac{7}{6}}$. You can also see that $b^{-4}=\frac{1}{b^4}$.

Put together, these become

$$\frac{a^{\frac{2}{3}}\cdot a^{\frac{1}{2}}\cdot b^2\cdot b^3}{a^{\frac{7}{6}}}\cdot \frac{1}{b^4}$$

You use the rules for multiplication and get

$$\frac{a^{\frac{2}{3}+\frac{1}{2}}\cdot b^{2+3}}{a^{\frac{7}{6}}}\cdot \frac{1}{b^4}$$

You write $\frac{2}{3}+\frac{1}{2}$ as a single fraction:

That gives you

$$\frac{a^{\frac{7}{6}}\cdot b^5}{a^{\frac{7}{6}}\cdot b^4}=\frac{\text{<msup><mrow><mi>a</mi></mrow><mrow><mfrac><mrow><mn>7</mn></mrow> <mrow><mn>6</mn></mrow></mfrac> </mrow></msup>}\cdot b^5}{\text{<msup><mrow><mi>a</mi></mrow><mrow><mfrac><mrow><mn>7</mn></mrow> <mrow><mn>6</mn></mrow></mfrac> </mrow></msup>}\cdot b^4}=b^{5-4}=b$$

Note! $b^1=b$.

Using the power rules you arrive at the following:

$$\begin{array}{llll}\hfill \frac{2^7\cdot \sqrt[3]{2}}{{\left(\sqrt[3]{2^5}\right)}^2}& =\frac{2^{7+\frac{1}{3}}}{2^{\frac{5}{3}\cdot 2}}& \hfill & \\ \hfill & =\frac{2^{\frac{21}{3}+\frac{1}{3}}}{2^{\frac{10}{3}}}& \hfill & \\ \hfill & =2^{\frac{22}{3}-\frac{10}{3}}& \hfill & \\ \hfill & =2^{\frac{12}{3}}& \hfill & \\ \hfill & =2^4& \hfill & \end{array}$$