Evaluate the expression \(\frac{b^2 + 2b + 4}{b^3 - 8}\) for \(b = \frac{1}{3}\).

Solution:

The expression is \( \frac{b^2 + 2b + 4}{b^3 - 8} \).

Factor the denominator as a difference of cubes: \(b^3 - 2^3 = (b-2)(b^2 + 2b + 4)\).

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

For \(b^2 + 2b + 4
eq 0\) (the discriminant is \(2^2 - 4(1)(4) = 4 - 16 = -12 < 0\), so this term is never zero), we can cancel out \(b^2 + 2b + 4\), simplifying the expression to \( \frac{1}{b-2} \).

Now, substitute \(b = \frac{1}{3}\):

\[ \frac{1}{\frac{1}{3} - 2} = \frac{1}{\frac{1}{3} - \frac{6}{3}} = \frac{1}{-\frac{5}{3}} = -\frac{3}{5} \]

Answer: -3/5