Question

(11-b2)- (3b-2b2)+(10-3b-b2)​

Answer 1

(11-b²)-(3b-2b²)+(10-3b-b²)=11-b²-3b+2b²+10-3b-b²=-6b+21

Answer 2

$(11 - {b}^{2} ) - (3b - 2 {b}^{2} ) + (10 - 3b - {b}^{2} ) = \\ 11 - {b}^{2} - 3b + 2 {b}^{2} + 10 - 3b - {b}^{2} = \\ 21 + 0 - 6b = 21 - 6b = 3 \times (7 - 2b)$