(x^2-1/x^2)^12

এর বিস্তৃতিতে x মুক্ত পদটি–

$\left(x^{2}+9+\frac{1}{x^{2}}\right)^{6}=\left(x+\frac{1}{x}\right)^{12}$

ধ্রুবক = 12-r-r = 0

r = 6

ধ্রুব পদ = ${ }^{12} e_{6}$ = $\frac{12 \times 1 \times 10 \times 8 \times 8 \times 7}{1 \times 7 \times 98 \times 4 \times 8 \times 8}$

$\begin{array}{l} =22 \times 42 \\ =924 \\ \end{array}$