Find the term independent of x in the expansion of :

To Find : term independent of $x$, i.e. $x^{0}$

For $\left(x-\frac{1}{x^{2}}\right)^{3 n}$

$a=x, b=-\frac{1}{x^{2}}$ and $N=3 n$

We have a formula

$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{c}\mathrm{N} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{N}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

$=\left(\begin{array}{c}3 n \\ r\end{array}\right)(x)^{3 n-r}\left(-\frac{1}{x^{2}}\right)^{r}$

$=\left(\begin{array}{c}3 n \\ r\end{array}\right)(x)^{3 n-r}(-1)^{r}\left(\frac{1}{x^{2}}\right)^{r}$

$=\left(\begin{array}{c}3 n \\ r\end{array}\right)(x)^{3 n-r}(-1)^{r}(x)^{-2 r}$

$=\left(\begin{array}{c}3 n \\ r\end{array}\right)(-1)^{r}(x)^{3 n-r-2 r}$

$=\left(\begin{array}{c}3 n \\ r\end{array}\right)(-1)^{r}(x)^{3 n-3 r}$

Now, to get coefficient of term independent of $x$ that is coefficient of $x^{0}$ we must have,

$(x)^{3 n-3 r}=x^{0}$

- $3 n-3 r=0$

- $3 r=3 n$

- $r=n$

Therefore, coefficient of $x^{0}=\left(\begin{array}{c}3 n \\ n\end{array}\right)(-1)^{n}$

$\underline{\text { Conclusion }}$ : coefficient of $x^{0}=\left(\begin{array}{c}3 n \\ n\end{array}\right)(-1)^{n}$