1. **State the problem:** Given the equations $x + y = 8$ and $x^2 + y^2 = 34$, we need to find: - $x^2 + 2xy + y^2$ - $2xy$ - $x^2 - 2xy + y^2$ - The values of $x$ and $y$ 2. From the first equation, square both sides: $$ (x + y)^2 = 8^2 $$ $$ x^2 + 2xy + y^2 = 64 $$ This gives us the value for $x^2 + 2xy + y^2$ as 64. 3. We know $x^2 + y^2 = 34$ from the second equation. Using the identity from step 2: $$ x^2 + 2xy + y^2 = 64 $$ $$ x^2 + y^2 + 2xy = 64 $$ Substitute $x^2 + y^2 = 34$: $$ 34 + 2xy = 64 $$ Solve for $2xy$: $$ 2xy = 64 - 34 = 30 $$ 4. Find $x^2 - 2xy + y^2$: Using the identity: $$ x^2 - 2xy + y^2 = (x - y)^2 $$ Rewrite: $$ x^2 + y^2 - 2xy $$ Substitute known values: $$ 34 - 30 = 4 $$ So, $x^2 - 2xy + y^2 = 4$. 5. Find the values of $x$ and $y$: We have: $$ (x + y)^2 = 64 $$ $$ (x - y)^2 = 4 $$ Take square roots: $$ x + y = 8 $$ $$ |x - y| = 2 $$ There are two cases: **Case 1:** $$ x - y = 2 $$ Add this to $x + y = 8$: $$ 2x = 10 \\ x = 5 $$ Find $y$: $$ 5 + y = 8 \\ y = 3 $$ **Case 2:** $$ x - y = -2 $$ Add to $x + y = 8$: $$ 2x = 6 \\ x = 3 $$ Find $y$: $$ 3 + y = 8 \\ y = 5 $$ So, the values of $(x,y)$ are $(5,3)$ or $(3,5)$. **Final answers:** - $x^2 + 2xy + y^2 = 64$ - $2xy = 30$ - $x^2 - 2xy + y^2 = 4$ - $x=5, y=3$ or $x=3, y=5$