1. **State the problem:** Simplify the expression $$\log_2 6 \cdot \log_3 6 - (\log_3 3 + \log_3 2)$$. 2. **Recall logarithm properties:** - $$\log_a a = 1$$ for any base $$a$$. - 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 base $$c$$. 3. **Evaluate terms inside parentheses:** - $$\log_3 3 = 1$$ because the base and argument are the same. - $$\log_3 2$$ remains as is. So, $$\log_3 3 + \log_3 2 = 1 + \log_3 2$$. 4. **Rewrite $$\log_2 6$$ and $$\log_3 6$$ using change of base to base 3:** - $$\log_2 6 = \frac{\log_3 6}{\log_3 2}$$. 5. **Substitute into the original expression:** $$\log_2 6 \cdot \log_3 6 - (1 + \log_3 2) = \frac{\log_3 6}{\log_3 2} \cdot \log_3 6 - 1 - \log_3 2$$ 6. **Simplify the product:** $$= \frac{(\log_3 6)^2}{\log_3 2} - 1 - \log_3 2$$ 7. **Express $$\log_3 6$$ as $$\log_3 (2 \cdot 3) = \log_3 2 + \log_3 3 = \log_3 2 + 1$$:** So, $$ (\log_3 6)^2 = (\log_3 2 + 1)^2 = (\log_3 2)^2 + 2 \log_3 2 + 1 $$ 8. **Substitute back:** $$= \frac{(\log_3 2)^2 + 2 \log_3 2 + 1}{\log_3 2} - 1 - \log_3 2$$ 9. **Split the fraction:** $$= \frac{(\log_3 2)^2}{\log_3 2} + \frac{2 \log_3 2}{\log_3 2} + \frac{1}{\log_3 2} - 1 - \log_3 2$$ 10. **Simplify terms:** $$= \log_3 2 + 2 + \frac{1}{\log_3 2} - 1 - \log_3 2$$ 11. **Cancel $$\log_3 2$$ and combine constants:** $$= 2 - 1 + \frac{1}{\log_3 2} = 1 + \frac{1}{\log_3 2}$$ 12. **Rewrite $$\frac{1}{\log_3 2}$$ using change of base:** $$\frac{1}{\log_3 2} = \log_2 3$$ 13. **Final simplified expression:** $$1 + \log_2 3$$ **Answer:** $$\boxed{1 + \log_2 3}$$