Given x ^ 2-4x + 4Y ^ 2-4y + 5 = 0, find the value of (x ^ 2 + y ^ 2) (x ^ 2-y ^ 2) - (x + y) ^ 2 (X-Y) ^ 2?

Given x ^ 2-4x + 4Y ^ 2-4y + 5 = 0, find the value of (x ^ 2 + y ^ 2) (x ^ 2-y ^ 2) - (x + y) ^ 2 (X-Y) ^ 2?

X ^ 2-4x + 4Y ^ 2-4y + 5 = 0 (x ^ 2-4x + 4) + (4Y ^ 2-4y + 1) = 0 (X-2) ^ 2 + (2y-1) ^ 2 = 0 X-2 = 02y-1 = 0 x = 2Y = 1 / 2 original formula = x ^ 4-y ^ 4 - [(x + y) (X-Y)] ^ 2 = x ^ 4-y ^ 4 - (x ^ 2-y ^ 2) ^ 2 = x ^ 4-y ^ 4 + 2x ^ 2Y ^ 2-y ^ 4 = 2 * 4 * 1 / 4-2 * 1 / 16 = 15 / 8