**Question:**

What must be subtracted from 3*a*2 − 6*ab* − 3*b*2 − 1 to get 4*a*2 − 7*ab* − 4*b*2 + 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 $=-a^{2}+a b+b^{2}-2$