Question

What must be subtracted from 3a² – 6ab – 3b² – 1 to get 4a² – 7ab – 4ab² + 1?

Answer 1

Let’s suppose the required number be x, 

So we have; 

(3a² – 6ab – 3b² – 1) – x = 4a² – 7a – 4b² + 1 

(3a² – 6ab – 3b² – 1) – (4a² – 7a – 4b2 + 1) = x 

So, 

To get the required number we have to subtract 4a² – 7a – 4b² + 1 from 3a² – 6ab – 3b² - 1

So, the required number is - a² + ab + b² – 2