1. **State the problem:** Simplify and evaluate the expression $$\left[\frac{3b^{-2}d \cdot 2 + (b \cdot d^{-1})^{2}}{12 b^{3} \cdot d^{-1}}\right]^{3}$$ 2. **Rewrite the numerator:** First term: $3b^{-2}d \cdot 2 = 6b^{-2}d$ Second term: $(b \cdot d^{-1})^{2} = b^{2} d^{-2}$ So numerator is $6b^{-2}d + b^{2} d^{-2}$ 3. **Rewrite the denominator:** $12 b^{3} d^{-1}$ 4. **Write the fraction:** $$\frac{6b^{-2}d + b^{2} d^{-2}}{12 b^{3} d^{-1}}$$ 5. **Split the fraction into two parts:** $$\frac{6b^{-2}d}{12 b^{3} d^{-1}} + \frac{b^{2} d^{-2}}{12 b^{3} d^{-1}}$$ 6. **Simplify each fraction separately:** First fraction: $$\frac{6b^{-2}d}{12 b^{3} d^{-1}} = \frac{6}{12} \cdot \frac{b^{-2}}{b^{3}} \cdot \frac{d}{d^{-1}} = \frac{1}{2} \cdot b^{-2-3} \cdot d^{1 - (-1)} = \frac{1}{2} b^{-5} d^{2}$$ Second fraction: $$\frac{b^{2} d^{-2}}{12 b^{3} d^{-1}} = \frac{1}{12} \cdot b^{2-3} \cdot d^{-2 - (-1)} = \frac{1}{12} b^{-1} d^{-1}$$ 7. **Combine the simplified terms:** $$\frac{1}{2} b^{-5} d^{2} + \frac{1}{12} b^{-1} d^{-1}$$ 8. **Find common denominator to combine if desired, or leave as is. Here, leave as is for clarity.** 9. **Now raise the entire expression to the power 3:** $$\left( \frac{1}{2} b^{-5} d^{2} + \frac{1}{12} b^{-1} d^{-1} \right)^{3}$$ This is the simplified form. Further expansion is possible but not requested. **Final answer:** $$\boxed{\left( \frac{1}{2} b^{-5} d^{2} + \frac{1}{12} b^{-1} d^{-1} \right)^{3}}$$