Evaluate the expression \(\frac{c^3 + 64}{3c^2 - 12c + 48}\) for \(c = 5\).

Solution:

The expression is \( \frac{c^3 + 64}{3c^2 - 12c + 48} \).

Factor the numerator as a sum of cubes: \(c^3 + 4^3 = (c+4)(c^2 - 4c + 16)\).

Factor out 3 from the denominator: \(3(c^2 - 4c + 16)\).

So, the expression becomes \( \frac{(c+4)(c^2 - 4c + 16)}{3(c^2 - 4c + 16)} \).

For \(c^2 - 4c + 16
eq 0\) (the discriminant is \((-4)^2 - 4(1)(16) = 16 - 64 = -48 < 0\), so this term is never zero), we can cancel out \(c^2 - 4c + 16\), simplifying the expression to \( \frac{c+4}{3} \).

Now, substitute \(c = 5\):

\[ \frac{5+4}{3} = \frac{9}{3} = 3 \]

Answer: 3