1. **State the problem:** We have the system of equations: $$\begin{cases} x + y = 6 \\ x^2 - y^2 = 2 \times 3! \end{cases}$$ We need to find the value of $xy$. 2. **Recall formulas and rules:** - The expression $x^2 - y^2$ can be factored as $(x+y)(x-y)$. - Factorials: $3! = 3 \times 2 \times 1 = 6$. 3. **Rewrite the second equation using factoring:** $$x^2 - y^2 = (x+y)(x-y)$$ Substitute $x+y=6$ and $3! = 6$: $$6 \times (x-y) = 2 \times 6$$ 4. **Simplify the right side:** $$6(x-y) = 12$$ 5. **Divide both sides by 6 to solve for $x-y$:** $$\cancel{6}(x-y) = \cancel{6} \times 2$$ $$x - y = 2$$ 6. **Now solve the system:** $$\begin{cases} x + y = 6 \\ x - y = 2 \end{cases}$$ Add the two equations: $$ (x+y) + (x-y) = 6 + 2 $$ $$ 2x = 8 $$ $$ x = 4 $$ 7. **Find $y$ by substituting $x=4$ into $x+y=6$:** $$4 + y = 6$$ $$y = 6 - 4 = 2$$ 8. **Calculate $xy$:** $$xy = 4 \times 2 = 8$$ **Final answer:** $$\boxed{8}$$