The constant term in the expansion of $(x^{2}-\frac{1}{x^{2}})^{16}$ is 

Let $T_{r+1}$ be the constant term in the expansion of $\left(x^{2}-\frac{1}{x^{2}}\right)^{16}$

$\begin{array}{l} \therefore \mathrm{T}_{\mathrm{r}+1}={ }^{16} \mathrm{C}_{r}\left(x^{2}\right)^{16-r}\left(\frac{-1}{x^{2}}\right)^{r} \\ ={ }^{16} C_{r}(x)^{32-2 r}(-1)^{r} \cdot \frac{1}{x^{2 r}} \\ =(-1)^{r} \cdot{ }^{16} C_{r}(x)^{32-2 r-2 r} \\ \Rightarrow(-1)^{r} \cdot{ }^{16} C_{r}(x)^{32-4 r} \end{array}$

For getting constant term, $32-4 r=0$

$\begin{array}{l} \Rightarrow r=8 \\ \therefore T_{r+1}=(-1)^{8} \cdot{ }^{16} C_{8}={ }^{16} C_{8} \end{array}$