Question

What must be subtracted from minus a cube b + 6 a square b square minus 7 a b + 4 to get minus 7 a square b square + 2 a cube b + 6 a b minus one?

Answer 1

Answer:

Step-by-step explanation:

Given What must be subtracted from minus a cube b + 6 a square b square minus 7 a b + 4 to get minus 7 a square b square + 2 a cube b + 6 a b minus one?

So given equation will be -a^3b + 6a^2b^2 – 7ab + 4 and – 7a^2b^2 + 2a^3b + 6ab – 1

For example what should be subtracted from 8 to get 3, so we get A – 8 = 3 or A = 3 + 8 = 11. So 11 – 8 = 3

In the same way we need to add the given equation

So  -a^3b + 6a^2b^2 – 7ab + 4 – A =   + 2a^3b – 7a^2b^2 + 6ab – 1

 -A = + 2a^3b – 7a^2b^2 + 6ab – 1 – (-a^3b + 6a^2b^2 – 7ab + 4)

 -A = 2a^3b – 7a^2b^2 + 6ab – 1 + a^3b - 6a^2b^2 + 7ab – 4

 A = 13a^2b^2 – 3a^3b – 13 ab + 5