1. **State the problem:** Given the equation $a x^3 - 27 = (2 x - 3)(4 x^2 + b x + 9)$, find the value of the product $a b$. 2. **Recall the formula and rules:** We will expand the right-hand side and then compare coefficients of powers of $x$ on both sides to find $a$ and $b$. 3. **Expand the right-hand side:** $$ (2 x - 3)(4 x^2 + b x + 9) = 2 x \cdot 4 x^2 + 2 x \cdot b x + 2 x \cdot 9 - 3 \cdot 4 x^2 - 3 \cdot b x - 3 \cdot 9 $$ Simplify each term: $$ = 8 x^3 + 2 b x^2 + 18 x - 12 x^2 - 3 b x - 27 $$ Group like terms: $$ = 8 x^3 + (2 b - 12) x^2 + (18 - 3 b) x - 27 $$ 4. **Compare with the left-hand side:** $$ a x^3 - 27 = 8 x^3 + (2 b - 12) x^2 + (18 - 3 b) x - 27 $$ Since the left side has no $x^2$ or $x$ terms, their coefficients must be zero: $$ 2 b - 12 = 0 \implies 2 b = 12 \implies b = 6 $$ $$ 18 - 3 b = 0 \implies 18 = 3 b \implies b = 6 $$ Both conditions agree that $b = 6$. 5. **Find $a$ by comparing $x^3$ coefficients:** $$ a = 8 $$ 6. **Calculate $a b$:** $$ a b = 8 \times 6 = 48 $$ **Final answer:** $a b = 48$. This corresponds to option (d).