2. \( \left(\frac{3}{4 a^{2}}-\frac{1}{3} a^{6}\right)^{3} \)

To simplify the expression \( \left(\frac{3}{4 a^{2}} - \frac{1}{3} a^{6}\right)^{3} \), we can start by rewriting it in a more manageable form. Let: \[ x = \frac{3}{4 a^{2}} - \frac{1}{3} a^{6} \] Now, we need to compute \( x^3 \). ### Step 1: Find a common denominator The common denominator for the two fractions inside the parentheses is \( 12a^2 \). We can rewrite each term: 1. For \( \frac{3}{4 a^{2}} \): \[ \frac{3}{4 a^{2}} = \frac{3 \cdot 3}{4 \cdot 3 a^{2}} = \frac{9}{12 a^{2}} \] 2. For \( -\frac{1}{3} a^{6} \): \[ -\frac{1}{3} a^{6} = -\frac{1 \cdot 4 a^{2}}{3 \cdot 4 a^{2}} = -\frac{4 a^{2}}{12 a^{2}} = -\frac{4 a^{2}}{12 a^{2}} \] ### Step 2: Combine the fractions Now we can combine the two fractions: \[ x = \frac{9}{12 a^{2}} - \frac{4 a^{2}}{12 a^{2}} = \frac{9 - 4 a^{4}}{12 a^{2}} \] ### Step 3: Raise to the power of 3 Now we need to compute \( x^3 \): \[ x^3 = \left(\frac{9 - 4 a^{4}}{12 a^{2}}\right)^3 = \frac{(9 - 4 a^{4})^3}{(12 a^{2})^3} \] ### Step 4: Simplify the denominator Calculating the denominator: \[ (12 a^{2})^3 = 12^3 a^{6} = 1728 a^{6} \] ### Step 5: Expand the numerator Now we need to expand \( (9 - 4 a^{4})^3 \) using the binomial theorem: \[ (9 - 4 a^{4})^3 = 9^3 - 3 \cdot 9^2 \cdot (4 a^{4}) + 3 \cdot 9 \cdot (4 a^{4})^2 - (4 a^{4})^3 \] Calculating each term: - \( 9^3 = 729 \) - \( 3 \cdot 9^2 \cdot (4 a^{4}) = 3 \cdot 81 \cdot 4 a^{4} = 972 a^{4} \) - \( 3 \cdot 9 \cdot (4 a^{4})^2 = 3 \cdot 9 \cdot 16 a^{8} = 432 a^{8} \) - \( (4 a^{4})^3 = 64 a^{12} \) Putting it all together: \[ (9 - 4 a^{4})^3 = 729 - 972 a^{4} + 432 a^{8} - 64 a^{12} \] ### Step 6: Combine everything Now we can write the final expression: \[ x^3 = \frac{729 - 972 a^{4} + 432 a^{8} - 64 a^{12}}{1728 a^{6}} \] ### Final Result Thus, the simplified expression is: \[ \left(\frac{3}{4 a^{2}} - \frac{1}{3} a^{6}\right)^{3} = \frac{729 - 972 a^{4} + 432 a^{8} - 64 a^{12}}{1728 a^{6}} \]