Mathematics - Binomial Theorem Question with Solution | TestHub

The term independent of$x$will come from

$\left(1-\frac{1}{x}+3x^5\right){\left(2x^2-\frac{1}{x}\right)}^8$

General term of${\left(2x^2-\frac{1}{x}\right)}^8$is${}^{8}C{}_r{\left(-\frac{1}{x}\right)}^r$${\left(2x^2\right)}^{8-r}=$${}^{8}C{}_r{2}^{8-r}{(-1)}^r{x}^{16-3r}$

From expansion

$1{\left(2x^2-\frac{1}{x}\right)}^8-\frac{1}{x}{\left(2x^2-\frac{1}{x}\right)}^8+3x^5{\left(2x^2-\frac{1}{x}\right)}^8$

$1{\left(2x^2-\frac{1}{x}\right)}^8\to$no constant term

$-\frac{1}{x}{\left(2x^2-\frac{1}{x}\right)}^8\to$for constant term

$r=5 \Rightarrow$$-{}^{8}C{}_3{2}^3{(-1)}^5=+448$

$3x^5{\left(2x^2-\frac{1}{x}\right)}^8\to$for constant term

$r=7\Rightarrow$${}^{38}C{}_7{2}^1{(-1)}^7=48$

Hence, term independent from$x=448-48=400$