a) \( \left(\frac{1}{2} a-\frac{1}{3} b\right)\left(\frac{1}{4} a^{2}-\frac{2}{3} a b+\frac{1}{2} b^{2}\right) \)

To expand the expression \( \left(\frac{1}{2} a - \frac{1}{3} b\right)\left(\frac{1}{4} a^{2} - \frac{2}{3} ab + \frac{1}{2} b^{2}\right) \), we can use the distributive property (also known as the FOIL method for binomials). This means we'll multiply each term in the first factor by each term in the second factor. 1. First, multiply \( \frac{1}{2} a \) by each term in the second expression: - \( \frac{1}{2} a \cdot \frac{1}{4} a^2 = \frac{1}{8} a^3 \) - \( \frac{1}{2} a \cdot \left(-\frac{2}{3} ab\right) = -\frac{1}{3} a^2 b \) - \( \frac{1}{2} a \cdot \frac{1}{2} b^2 = \frac{1}{4} ab^2 \) 2. Next, multiply \( -\frac{1}{3} b \) by each term in the second expression: - \( -\frac{1}{3} b \cdot \frac{1}{4} a^2 = -\frac{1}{12} a^2 b \) - \( -\frac{1}{3} b \cdot \left(-\frac{2}{3} ab\right) = \frac{2}{9} ab^2 \) - \( -\frac{1}{3} b \cdot \frac{1}{2} b^2 = -\frac{1}{6} b^3 \) 3. Now, combine all these results: \[ \frac{1}{8} a^3 + \left(-\frac{1}{3} a^2 b - \frac{1}{12} a^2 b\right) + \left(\frac{1}{4} ab^2 + \frac{2}{9} ab^2\right) - \frac{1}{6} b^3 \] 4. Combine like terms: - For \( a^2b \): \[ -\frac{1}{3} a^2 b - \frac{1}{12} a^2 b = -\frac{4}{12} a^2 b - \frac{1}{12} a^2 b = -\frac{5}{12} a^2 b \] - For \( ab^2 \): \[ \frac{1}{4} ab^2 + \frac{2}{9} ab^2 = \left(\frac{9}{36} + \frac{8}{36}\right)ab^2 = \frac{17}{36} ab^2 \] Finally, we can write the expanded expression as: \[ \frac{1}{8} a^3 - \frac{5}{12} a^2 b + \frac{17}{36} ab^2 - \frac{1}{6} b^3 \]