Question

2. What should be subtracted from 2a2 - b2 + 3a²b to get a2 + b2 + 3ab2 - 6?​

Answer 1

Answer:

$(2 {a}^{2} - {b}^{2} + 3 {a}^{2} b) + ( {a}^{2} + {b}^{2} + 3a {b}^{2} - 6) \\ \\ = 2 {a}^{2} - {b}^{2} + 3 {a}^{2} b + {a}^{2} + {b}^{2} + 3a {b}^{2} - 6 \\ \\ = 3 {a}^{2} + 3 {a}^{2} b+ 3a {b}^{2} - 6$

according to me 3a^2 + 3a^2b + 3ab^2 - 6 must be subtracted

Answer 2

Answer:

(2a

2

−b

2

+3a

2

b)+(a

2

+b

2

+3ab

2

−6)

=2a

2

−b

2

+3a

2

b+a

2

+b

2

+3ab

2

−6

=3a

2

+3a

2

b+3ab

2

−6