What must be subtracted from $3 a^{2} - 6 a b - 3 b^{2} - 1$ to get $4 a^{2} - 7 a b - 4 b^{2}$ .

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Solution

Let x be subtracted.
Then, $(3 a^{2} - 6 a b - 3 b^{2} - 1) - x = 4 a^{2} - 7 a b - 4 b^{2}$
$\Rightarrow 3 a^{2} - 6 a b - 3 b^{2} - 1 - 4 a^{2} + 7 a b + 4 b^{2} = x$
$\Rightarrow 3 a^{2} - 4 a^{2} - 6 a b + 7 a b - 3 b^{2} + 4 b^{2} - 1 = x$
$\Rightarrow∴ x = a b - a^{2} + b^{2} - 1$ .

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