1. Stating the problem: Solve the equation $ (x+1)^2 + (x-2)^2 = (x+2)^2 + (x-1)^2 $. 2. Formula and rules: Use the expansion formula for squares: $ (a+b)^2 = a^2 + 2ab + b^2 $. 3. Expand each term: $$ (x+1)^2 = x^2 + 2x + 1 $$ $$ (x-2)^2 = x^2 - 4x + 4 $$ $$ (x+2)^2 = x^2 + 4x + 4 $$ $$ (x-1)^2 = x^2 - 2x + 1 $$ 4. Substitute expansions into the equation: $$ (x^2 + 2x + 1) + (x^2 - 4x + 4) = (x^2 + 4x + 4) + (x^2 - 2x + 1) $$ 5. Combine like terms on each side: Left side: $ x^2 + x^2 + 2x - 4x + 1 + 4 = 2x^2 - 2x + 5 $ Right side: $ x^2 + x^2 + 4x - 2x + 4 + 1 = 2x^2 + 2x + 5 $ 6. Set the equation: $$ 2x^2 - 2x + 5 = 2x^2 + 2x + 5 $$ 7. Subtract $2x^2$ and 5 from both sides: $$ \cancel{2x^2} - 2x + \cancel{5} - \cancel{2x^2} = \cancel{2x^2} + 2x + \cancel{5} - \cancel{2x^2} - \cancel{5} $$ Simplifies to: $$ -2x = 2x $$ 8. Add $2x$ to both sides: $$ -2x + 2x = 2x + 2x $$ $$ 0 = 4x $$ 9. Divide both sides by 4: $$ \frac{0}{\cancel{4}} = \frac{4x}{\cancel{4}} $$ $$ 0 = x $$ 10. Final answer: $ x = 0 $.