Question

plz solve qs no.4 without expanding

Answer 1

|a+b+2c a b |
| c b+c+2a b |
| c a c+a+2b|

=
|2 (a+b+c) a b |
|2 (a+b+c) b+c+2a b |
|2 (a+b+c) a c+a+2b|
........ [ C1' = C1 +C2+C3]
=
2 (a+b+c)| 1 a b |
| 1 b+c+2a b|
| 1 a c+a+2b|
... [ taking common 2 (a+b+2c) from C1 ]
=
2( a+b+c) | 1 a b |
|0 a+b+c 0 |
| 0 0 a+b+c|
....... [ R2'=R2 -R1
R3'=R3-R1 ]

=
2(a+b+c)×(a+b+c)2 (square )
...[ expanding the determinant in respect of R1]
=2 ( a+b+c)3 (qube) PROVED