Prove that
$x - (a + b)^{2} + y - (b - a)^{2} = x - (a - b)^{2} + y - (a + b)^{2}$

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Solution

$x - (a + b)^{2} + y - (b - a)^{2}$

$x - (a^{2} + b^{2} + 2 a b) + y - (b^{2} + a^{2} - 2 a b)$

$x - 2 a^{2} - 2 b^{2} + y$ ...(1)

$x - (a - b)^{2} + y - (b + a)^{2}$

$x - (a^{2} + b^{2} - 2 a b) + y - (b^{2} + a^{2} + 2 a b)$

$x - 2 a^{2} - 2 b^{2} + y$ ...(2)

From (1) and (2) it's clear that:

$x - (a + b)^{2} + y - (b - a)^{2} = x - (a - b)^{2} + y - (b + a)^{2}$

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