In the expansion of `(x^(2) - (1)/(x^(2)))^(16)`, the value of constant term is......

Explanation

To find the constant term in the expansion of \((x^2 - \frac{1}{x^2})^{16}\), we can follow these steps:

### Step 1: Identify the General Term
The general term \(T_r\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = x^2\), \(b = -\frac{1}{x^2}\), and \(n = 16\). Thus, the general term is:
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