1. **State the problem:** Simplify the expression $$\log_2 6 \cdot \log_3 6 - (\log_2 3 + \log_3 2)$$. 2. **Recall the change of base formula:** For any positive numbers $a,b,c$ with $a \neq 1$, $$\log_a b = \frac{\log_c b}{\log_c a}$$. 3. **Express all logarithms in terms of natural logs (or any common base):** $$\log_2 6 = \frac{\ln 6}{\ln 2}, \quad \log_3 6 = \frac{\ln 6}{\ln 3}, \quad \log_2 3 = \frac{\ln 3}{\ln 2}, \quad \log_3 2 = \frac{\ln 2}{\ln 3}$$ 4. **Rewrite the expression:** $$\left(\frac{\ln 6}{\ln 2} \cdot \frac{\ln 6}{\ln 3}\right) - \left(\frac{\ln 3}{\ln 2} + \frac{\ln 2}{\ln 3}\right) = \frac{(\ln 6)^2}{\ln 2 \cdot \ln 3} - \left(\frac{\ln 3}{\ln 2} + \frac{\ln 2}{\ln 3}\right)$$ 5. **Find a common denominator for the subtraction:** $$\frac{(\ln 6)^2}{\ln 2 \cdot \ln 3} - \frac{(\ln 3)^2}{\ln 2 \cdot \ln 3} - \frac{(\ln 2)^2}{\ln 2 \cdot \ln 3} = \frac{(\ln 6)^2 - (\ln 3)^2 - (\ln 2)^2}{\ln 2 \cdot \ln 3}$$ 6. **Use the property of logarithms:** $$\ln 6 = \ln (2 \cdot 3) = \ln 2 + \ln 3$$ 7. **Expand the numerator:** $$ (\ln 6)^2 - (\ln 3)^2 - (\ln 2)^2 = (\ln 2 + \ln 3)^2 - (\ln 3)^2 - (\ln 2)^2 $$ $$= (\ln 2)^2 + 2 \ln 2 \ln 3 + (\ln 3)^2 - (\ln 3)^2 - (\ln 2)^2 = 2 \ln 2 \ln 3$$ 8. **Substitute back:** $$\frac{2 \ln 2 \ln 3}{\ln 2 \cdot \ln 3}$$ 9. **Cancel common factors:** $$\frac{\cancel{2} \cancel{\ln 2} \cancel{\ln 3}}{\cancel{\ln 2} \cdot \cancel{\ln 3}} = 2$$ **Final answer:** $$2$$