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

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

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

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

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$b^{2}$

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Solution

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

$(a b + 1)^{2}$ - $(a b)^{2}$
$\Rightarrow (a b + 1)^{2} - (a b)^{2} = (a (a + 1) + 1)^{2} - (a (a + 1))^{2}$ [ $∵$ b = 1 + a]
$= a^{2} \times ((a + 1)^{2} + 1 + 2 a (a + 1)) - a^{2} \times (a + 1)^{2}$
$= 2 a^{2} + 2 a + 1$
$= a^{2} + 2 a + 1 + a^{2}$
$= (a + 1)^{2} + a^{2}$
$= b^{2} + a^{2}$

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