What must be subtracted from 3a 2 – 6ab – 3b 2 – 1 to get 4a 2 – 7ab – 4ab 2 + 1?

Question

What must be subtracted from 3a 2 &#x2013; 6ab &#x2013; 3b 2 &#x2013; 1 to get 4a 2 &#x2013; 7ab &#x2013; 4ab 2 + 1?

Philoid · Accepted Answer

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