2.3) \( \frac{p^{3}+q^{3}}{6 p^{2}} \times \frac{3 p-3 q}{p^{2}-q^{2}} \)

To simplify the expression, we can start with the first fraction \( \frac{p^{3} + q^{3}}{6 p^{2}} \). We know that \( p^{3} + q^{3} \) can be factored using the sum of cubes formula: \[ p^{3} + q^{3} = (p + q)(p^{2} - pq + q^{2}). \] Therefore, we rewrite the first fraction: \[ \frac{(p + q)(p^{2} - pq + q^{2})}{6 p^{2}}. \] Now, moving on to the second part \( \frac{3p - 3q}{p^{2} - q^{2}} \), we can factor out a 3 from the numerator and also recognize that the denominator is a difference of squares: \[ p^{2} - q^{2} = (p - q)(p + q). \] Thus, we rewrite the second fraction as: \[ \frac{3(p - q)}{(p - q)(p + q)}. \] Assuming \( p \neq q \) to avoid division by zero, we can simplify this to: \[ \frac{3}{p + q}. \] Now, combining our simplified fractions, we have: \[ \frac{(p + q)(p^{2} - pq + q^{2})}{6 p^{2}} \times \frac{3}{p + q}. \] The \( p + q \) terms cancel out, leaving us with: \[ \frac{(p^{2} - pq + q^{2})}{2 p^{2}}. \] So the final simplified form of the expression is: \[ \frac{p^{2} - pq + q^{2}}{2 p^{2}}. \]