Question

subtract the sum of 9b2- 3c2 and 2b2+bc-2c2 from the sum of 2b2-2bc-c2 and c2+ 2bc- b2​

Answer 1

Answer:

$9 {b}^{2} - 3 {c}^{2} + 2 {b}^{2} + bc - 2 {c }^{2} = 9 {b}^{2} + 2 {b}^{2} - 3 {c}^{2} - 2 {c}^{2} + bc \: = 11 {b}^{2} - 5 {c}^{2} + bc$

$2 {b}^{2} - 2bc - {c}^{2} + {c}^{2} + 2 - {b}^{2} = 2 {b}^{2} - {b}^{2} - 2bc + 0 = {b}^{2} - 2bc$

now,

$({b}^{2} - 2bc) - (11 {b}^{2} - 5 {c}^{2} + bc ) \: = {b}^{2} - \: 2bc - 11 {b}^{2} + 5 {c}^{2} - bc \: = {b}^{2} - 11 {b}^{2} - 2bc - bc + 5 {c}^{2} \: = - 10 {b}^{2} - 3bc + 5 {c}^{2}$

Answer 2

Answer:

9b

2

−3c

2

+2b

2

+bc−2c

2

=9b

2

+2b

2

−3c

2

−2c

2

+bc=11b

2

−5c

2

+bc

2 {b}^{2} - 2bc - {c}^{2} + {c}^{2} + 2 - {b}^{2} = 2 {b}^{2} - {b}^{2} - 2bc + 0 = {b}^{2} - 2bc2b

2

−2bc−c

2

+c

2

+2−b

2

=2b

2

−b

2

−2bc+0=b

2

−2bc

now,

({b}^{2} - 2bc) - (11 {b}^{2} - 5 {c}^{2} + bc ) \: = {b}^{2} - \: 2bc - 11 {b}^{2} + 5 {c}^{2} - bc \: = {b}^{2} - 11 {b}^{2} - 2bc - bc + 5 {c}^{2} \: = - 10 {b}^{2} - 3bc + 5 {c}^{2}(b

2

−2bc)−(11b

2

−5c

2

+bc)=b

2

−2bc−11b

2

+5c

2

−bc=b

2

−11b

2

−2bc−bc+5c

2

=