Solve $(a b + 1)^{2}$ - $(a b)^{2}$ where, b - a = 1.

$a^{2}$ + $(b + 1)^{2}$

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$b^{2}$

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$(a + 1)^{2}$ + $(b + 1)^{2}$

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$a^{2}$ + $b^{2}$

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Solution

The correct option is D
$a^{2}$ + $b^{2}$

$(a b + 1)^{2} - (a b)^{2} = (a (a + 1) + 1)^{2} - (a (a + 1))^{2}$
$= (a^{2} (a + 1)^{2} + 1 + 2 a (a + 1)) - (a^{2} (a + 1)^{2})$
$= 2 a (a + 1) + 1$
$= 2 a^{2} + 2 a + 1$
$= a^{2} + 2 a + 1 + a^{2}$
$= (a + 1)^{2} + a^{2}$
$= b^{2} + a^{2}$ [since $b = a + 1$ ]

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