1. **State the problem:** We are given three products: $A \times B = 36$, $B \times C = 18$, and $A \times C = 12$. We need to find the value of $A \times B \times C$. 2. **Recall the formula:** To find $A \times B \times C$, we can use the given products and the property of multiplication. 3. **Multiply all three given equations:** $$ (A \times B) \times (B \times C) \times (A \times C) = 36 \times 18 \times 12 $$ 4. **Rewrite the left side:** $$ (A \times B) \times (B \times C) \times (A \times C) = A^2 \times B^2 \times C^2 $$ 5. **So:** $$ A^2 \times B^2 \times C^2 = 36 \times 18 \times 12 $$ 6. **Calculate the right side:** $$ 36 \times 18 = 648 $$ $$ 648 \times 12 = 7776 $$ 7. **Therefore:** $$ A^2 \times B^2 \times C^2 = 7776 $$ 8. **Take the square root of both sides:** $$ A \times B \times C = \sqrt{7776} $$ 9. **Simplify $\sqrt{7776}$:** Prime factorization of 7776: $$ 7776 = 2^5 \times 3^5 $$ So, $$ \sqrt{7776} = \sqrt{2^5 \times 3^5} = 2^{\frac{5}{2}} \times 3^{\frac{5}{2}} = 2^2 \times 3^2 \times \sqrt{2 \times 3} = 4 \times 9 \times \sqrt{6} = 36 \sqrt{6} $$ 10. **Final answer:** $$ A \times B \times C = 36 \sqrt{6} $$