Question

Find the term independent of $x$ in the expansion of ${[\frac{1}{2}{x}^{\frac{1}{3}}+x^{-\frac{1}{5}}]}^8$

Answer 1

Writing the general term, we get
$T_{r+1}={}^8{C}_r{x}^{\frac{8-r}{3}-\frac{r}{5}}.\frac{1}{2^{8-r}}$
$={}^8{C}_r{x}^{\frac{40-8r}{15}}.\frac{1}{2^{8-r}}$
Hence for term independent of $x$.
$40-8r=0$
$r=5$
Hence
$T_6$ is independent of $x$.
$={}^8{C}_5.\frac{1}{8}$
$=\frac{56}{8}$
$=\frac{7}{2}$.