In the expansion of

(b) $T_{6}$

Suppose the (r + 1)th term in the given expansion is independent of x.

Thus, we have:

$T_{r+1}={ }^{8} C_{r}\left(\frac{1}{2} x^{1 / 3}\right)^{8-r}\left(x^{-1 / 5}\right)^{r}$

$={ }^{8} C_{r} \frac{1}{2^{8-r}} x^{\frac{8-r}{3}-\frac{r}{5}}$

For this term to be independent of $x$, we must have

$\frac{8-r}{3}-\frac{r}{5}=0$

$\Rightarrow 40-5 r-3 r=0$

$\Rightarrow r=5$

Hence, the required term is the 6 th term, i.e. $T_{6}$