The term independent of x in the expansion of $$\left(3x^{3} - \frac{4}{x}\right)^8$$

let  $$(r+1)^{th}$$ term be the independent of x which is given by bio nominal theorem 

$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$  $$a^{(8-r)}$$ $$\times$$ $$b^{r}$$ 

here we have a = 3$$x^{3}$$ and b = -4\x 

put the value we get 

$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$  (3$$x^{3}$$)^(8-r)  $$\times$$ (-4\x)^r

$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$ (3)^(8-r)($$x^{3}$$)^(8-r) $$\times$$ (-4)^r(x)^(-r)

$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$ (3)^(8-r)($$x^{24-3r}$$) $$\times$$ (-4)^r(x)^(-r)

since  the term is independent of x, we have 

24 - 4r = 0

r  =  6 

so the (r+1) = 7 term is independent so we can say that 

$$T_{7}$$ = 8$$c_{6}$$ $$\times$$ $$3^{3-1}$$ $$\times$$ (-4)^2(6-1)

$$T_{7}$$ = 8 $$\times$$ 7 $$\times$$  $$3^{2}$$ $$\times$$ $$2^{11}$$

$$T_{7}$$ = $$2^{14}$$ $$\times$$ 7 $$\times$$ $$3^{2}$$  Answer