Explanation:
we just do the multiplications that squaring is :
(x² + y²)² = (x² + y²)(x² + y²) =
= x⁴ + x²y² + x²y² + y⁴ = x⁴ + 2x²y² + y⁴
(x² - y²)² + (2xy)² = (x² - y²)(x² - y²) + 2xy×2xy =
= x⁴ - x²y² - x²y² + y⁴ + 4x²y² =
= x⁴ - 2x²y² + y⁴ + 4x²y² =
= x⁴ + 2x²y² + y⁴
as you can see, both results are the same.
therefore, the original equation is a true identity.