1. **State the problem:** Simplify the expression $$\log_2 6 \cdot \log_3 6 - (\log_3 3 + \log_3 2)$$. 2. **Recall logarithm properties:** - Product rule: $$\log_a (xy) = \log_a x + \log_a y$$ - Change of base formula: $$\log_a b = \frac{\log_c b}{\log_c a}$$ for any positive base $c \neq 1$ - $$\log_a a = 1$$ 3. **Simplify inside the parentheses:** $$\log_3 3 + \log_3 2 = 1 + \log_3 2$$ 4. **Rewrite the expression:** $$\log_2 6 \cdot \log_3 6 - (1 + \log_3 2)$$ 5. **Express $\log_2 6$ and $\log_3 6$ using change of base to natural logs (or common logs):** $$\log_2 6 = \frac{\log 6}{\log 2}, \quad \log_3 6 = \frac{\log 6}{\log 3}$$ 6. **Calculate the product:** $$\log_2 6 \cdot \log_3 6 = \frac{\log 6}{\log 2} \cdot \frac{\log 6}{\log 3} = \frac{(\log 6)^2}{\log 2 \cdot \log 3}$$ 7. **Rewrite the entire expression:** $$\frac{(\log 6)^2}{\log 2 \cdot \log 3} - 1 - \log_3 2$$ 8. **Express $\log_3 2$ using change of base:** $$\log_3 2 = \frac{\log 2}{\log 3}$$ 9. **Substitute back:** $$\frac{(\log 6)^2}{\log 2 \cdot \log 3} - 1 - \frac{\log 2}{\log 3}$$ 10. **Rewrite $\log 6$ as $\log (2 \cdot 3) = \log 2 + \log 3$:** $$\frac{(\log 2 + \log 3)^2}{\log 2 \cdot \log 3} - 1 - \frac{\log 2}{\log 3}$$ 11. **Expand numerator:** $$\frac{(\log 2)^2 + 2 \log 2 \log 3 + (\log 3)^2}{\log 2 \cdot \log 3} - 1 - \frac{\log 2}{\log 3}$$ 12. **Split the fraction:** $$\frac{(\log 2)^2}{\log 2 \log 3} + \frac{2 \log 2 \log 3}{\log 2 \log 3} + \frac{(\log 3)^2}{\log 2 \log 3} - 1 - \frac{\log 2}{\log 3}$$ 13. **Simplify terms:** $$\frac{\log 2}{\log 3} + 2 + \frac{\log 3}{\log 2} - 1 - \frac{\log 2}{\log 3}$$ 14. **Cancel $\frac{\log 2}{\log 3}$ terms:** $$2 + \frac{\log 3}{\log 2} - 1 = 1 + \frac{\log 3}{\log 2}$$ 15. **Rewrite $\frac{\log 3}{\log 2}$ as $\log_2 3$:** $$1 + \log_2 3$$ **Final answer:** $$\boxed{1 + \log_2 3}$$