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

To expand \( \left(\frac{2 a}{3}-\frac{b^{2}}{4}\right)^{3} \), we can use the binomial theorem, which states that \( (x - y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} (-y)^k \). Here, let \( x = \frac{2a}{3} \) and \( y = \frac{b^2}{4} \), and \( n = 3 \). Now, we calculate the expansion: \[ \left(\frac{2 a}{3}-\frac{b^{2}}{4}\right)^{3} = \sum_{k=0}^{3} \binom{3}{k} \left(\frac{2 a}{3}\right)^{3-k} \left(-\frac{b^{2}}{4}\right)^{k} \] Calculating for each \( k \) from 0 to 3: 1. **For \( k = 0 \)**: \[ \binom{3}{0} \left(\frac{2a}{3}\right)^{3} = 1 \cdot \left(\frac{8a^{3}}{27}\right) = \frac{8a^{3}}{27} \] 2. **For \( k = 1 \)**: \[ \binom{3}{1} \left(\frac{2 a}{3}\right)^{2} \left(-\frac{b^{2}}{4}\right) = 3 \cdot \left(\frac{4a^{2}}{9}\right) \cdot \left(-\frac{b^{2}}{4}\right) = -\frac{3a^{2}b^{2}}{12} = -\frac{a^{2}b^{2}}{4} \] 3. **For \( k = 2 \)**: \[ \binom{3}{2} \left(\frac{2 a}{3}\right) \left(-\frac{b^{2}}{4}\right)^{2} = 3 \cdot \left(\frac{2 a}{3}\right) \cdot \left(\frac{b^{4}}{16}\right) = \frac{6ab^{4}}{48} = \frac{ab^{4}}{8} \] 4. **For \( k = 3 \)**: \[ \binom{3}{3} \left(-\frac{b^{2}}{4}\right)^{3} = 1 \cdot \left(-\frac{b^{6}}{64}\right) = -\frac{b^{6}}{64} \] Now, summing all these results together: \[ \frac{8 a^{3}}{27} - \frac{a^{2}b^{2}}{4} + \frac{ab^{4}}{8} - \frac{b^{6}}{64} \] Thus, the expansion of \( \left(\frac{2 a}{3}-\frac{b^{2}}{4}\right)^{3} \) is: \[ \frac{8 a^{3}}{27} - \frac{a^{2}b^{2}}{4} + \frac{ab^{4}}{8} - \frac{b^{6}}{64} \]