Strategy: For this question, we can either expand \((\frac{1}{4-\sqrt{15}})^2\) and then "fix" the resulting denominator, or " fix" the denominator first and then expand.
Let's fix the denominator first

Take: \(\frac{1}{4-\sqrt{15}}\)

Multiply numerator and denominator by \(4+\sqrt{15}\), which is the complement of \(4-\sqrt{15}\)

When we do this we get: \(\frac{1}{4-\sqrt{15}} = \frac{(1)(4+\sqrt{15})}{(4-\sqrt{15})(4+\sqrt{15})}= \frac{4+\sqrt{15}}{16 - 15}= \frac{4+\sqrt{15}}{1} = 4+\sqrt{15}\)

Substitute this "fixed" expression into the original expression to get: \((\frac{1}{4-\sqrt{15}})^2 = (4+\sqrt{15})^2 = (4+\sqrt{15})(4+\sqrt{15}) = 16 + 4\sqrt{15} + 4\sqrt{15} + (\sqrt{15})^2 = 31 + 8\sqrt{15}\)

Answer: C