$\left(x^{3}-3 x+\frac{3}{x}-\frac{1}{x^{3}}\right)^{2}$ এর বিস্তৃতিতে x বর্জিত পদটি কত?

$\left(x^{3}-3 x+\frac{3}{x}-\frac{1}{x^{3}}\right)^{2}=\left(x^{3}-\frac{1}{x^{3}}-3 x+\frac{3}{x}\right)^{2}$ $=\left\{x^{3}-\left(\frac{1}{x}\right)^{3}-3 \cdot x^{2} \cdot \frac{1}{x}+3 \cdot x \cdot\left(\frac{1}{x}\right)^{2}\right\}^{2}=\left\{\left(x-\frac{1}{x}\right)^{3}\right\}^{2}=\left(x-\frac{1}{x}\right)^{6}$ $T_{r+1}={ }^{6} C_{r} x^{6-r}(-1)^{r} x^{-r}={ }^{6} C_{r}(-1)^{r} x^{6-2 r}$ এখন, $6-2 r=0 \Rightarrow 2 r=6 \Rightarrow r=3$ $\therefore \mathrm{x}$ বৰ্জিত পদ $=(-1)^{3}{ }^{6} C_{3}=-20$