Find the value of (\(\frac{1}{6})^{-2} +( \frac{1}{5})^{-2} + (\frac{1}{4})^{-2}\)

Expression: $ \left(\frac{1}{6}\right)^{-2} + \left(\frac{1}{5}\right)^{-2} + \left(\frac{1}{4}\right)^{-2} $
$ \text{Step 1: Use the rule } (a/b)^{-n} = (b/a)^n $
$ \left(\frac{1}{6}\right)^{-2} = 6^2 = 36 $
$ \left(\frac{1}{5}\right)^{-2} = 5^2 = 25 $
$ \left(\frac{1}{4}\right)^{-2} = 4^2 = 16 $
$ \text{Step 2: Add the values } $
$ 36 + 25 + 16 = 77 $
$ \therefore \left(\frac{1}{6}\right)^{-2} + \left(\frac{1}{5}\right)^{-2} + \left(\frac{1}{4}\right)^{-2} = 77 $