Question:
What must be subtracted from 3a2 − 6ab − 3b2 − 1 to get 4a2 − 7ab − 4b2 + 1?
Solution:
Let the required number be
$\left(3 a^{2}-6 a b-3 b^{2}-1\right)-x=4 a^{2}-7 a b-4 b^{2}+1$
$\left(3 a^{2}-6 a b-3 b^{2}-1\right)-\left(4 a^{2}-7 a b-4 b^{2}+1\right)=x$
$\therefore$ Required number $=-a^{2}+a b+b^{2}-2$
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