Thank you for reporting, we will resolve it shortly
Q.
The term independent of $x$ in the binomial expansion of $\left(1-\frac{1}{x}+3 x^5\right)\left(2 x^2-\frac{1}{x}\right)^8$ is
Binomial Theorem
Solution:
Given expression $=\left(1-\frac{1}{x}+3 x^5\right)\left(2 x^2-\frac{1}{x}\right)^8$
General term in $\left(2 x^2-\frac{1}{x}\right)^8={ }^8 C_r\left(2 x^2\right)^{8-r}\left(\frac{-1}{x}\right)^r$
$\therefore 16-3 r =0 \Rightarrow r =\frac{16}{3}$ which is not possible
and $ 16-3 r =1 \Rightarrow r =5$
and $ 16-3 r =-5 \Rightarrow r =7$
$\therefore$ Term independent of $x =-{ }^8 C _5 2^3(-1)^5+3 \cdot{ }^8 C _7 \cdot 2(-1)^7$
$=56 \times 8-48 $
$=400 $