Right option is (c) (Σab)^3
For explanation: Applying R1 –> R1 + R2 + R3
\(\begin{vmatrix}\Sigma ab & \Sigma ab & \Sigma ab \\ab + bc & -ca & ab + bc \\bc + ca & bc + ca & -ab \end {vmatrix}\)
This is equal to,
= Σab \(\begin{vmatrix}1 & 1 & 1 \\ab + bc & -ca & ab + bc \\bc + ca & bc + ca & -ab \end {vmatrix}\)
Applying C1 –> C1 – C2 and C2 –> C2 – C3
= Σab \(\begin{vmatrix}0 & 0 & 1 \\ \Sigma ab & -\Sigma ab & ab + bc \\0 & \Sigma ca & -ab \end {vmatrix}\)
= (Σab)^3