Explanation

(a)
Last term of $\left( 2^{\frac{1}{3}} - \frac{1}{\sqrt{2\ }} \right)^{n}$ is
$T_{n + 1} = \ ^{n}C_{n}\left( 2^{\frac{1}{3}} \right)^{n - n}\left( - \frac{1}{\sqrt{2}} \right)^{n}$
$= \ ^{n}C_{n}( - 1)^{n}\frac{1}{2^{\frac{n}{2}}} = \frac{( - 1)^{n}}{2^{\frac{n}{2}}}$
Also, we have
$\left( \frac{1}{3^{\frac{5}{3}}} \right)^{\log_{3}8} = 3^{- (\frac{5}{3})\log_{3}2^{3}} = 2^{- 5}$
Thus, $\frac{{( - 1)}^{n}}{2^{\frac{n}{2}}} = 2^{- 5}\ \Rightarrow \ \frac{{( - 1)}^{n}}{2^{\frac{n}{2}}} = \frac{{( - 1)}^{10}}{2^{5}}$
$\Rightarrow \ \frac{n}{2} = 5\ \Rightarrow \ n = 10$
Now, $T_{5} = T_{4 + 1} = \ ^{10}C_{4}{(2^{\frac{1}{3}})}^{10 - 4}\left( - \frac{1}{\sqrt{2}} \right)^{4}$
$= \frac{10!}{4!6!}{(2^{\frac{1}{3}})}^{6}{( - 1)}^{4}{(2^{\frac{- 1}{2}})}^{4}$
$= 210(2)^{2}(1)\left( 2^{- 2} \right) = 210$