Question

In the expansion of ${\left(x^2-\frac{1}{3x}\right)}^9$, the term without x is equal to
(a) $\frac{28}{81}$

(b) $\frac{-28}{243}$

(c) $\frac{28}{243}$

(d) none of these

Answer 1

(c) $\frac{28}{243}$

Suppose the (r + 1)th term in the given expansion is independent of x.
Then , we have:
$$\begin{gathered} T_{r+1}={}^{9}C_r(x^2{)}^{9-r}{\left(\frac{-1}{3x}\right)}^r \\ =(-1{)}^r{}^{9}C_r\frac{1}{3^r}{x}^{18-2r-r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have}: \\ 18-3r=0 \\ \Rightarrow r=6 \\ \therefore \mathrm{Required}\mathrm{term}=(-1{)}^6{}^{9}C_6\frac{1}{3^6}=\frac{9\times 8\times 7}{3\times 2}\times \frac{1}{3^6}=\frac{28}{243} \end{gathered}$$