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

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Solution

$3 a^{2} - 6 a b - 3 b^{2} - 1 4 a^{2} - 7 a b - 4 b^{2} + 1 - + + - - -------------------- - a^{2} + a b + b^{2} - 2 - ------------------- -$

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