Right option is (d) –(a^3 + b^3 + c^3 – 3abc)
The explanation: Given, \(\begin{vmatrix}a & b & c \\b & c & a \\c & a & b \end {vmatrix}\)
Replacing R1 = R1 + R2 + R3
\(\begin{vmatrix}a + b + c & a + b + c & a + b + c \\b & c & a \\c & a & b \end {vmatrix}\)
= (a + b + c)\(\begin{vmatrix}1 & 1 & 1 \\b & c & a \\c & a & b \end {vmatrix}\)
Replacing 2^nd column by C2 – C1 and 3^rd column by C3 – C1
= (a + b + c)\(\begin{vmatrix}1 & 0 & 0 \\b & c-b & a-b \\c & a-c & b-c \end {vmatrix}\)
= (a + b + c)[(c – b)(b – c) – (a – b)(a – c)]
= (a + b + c)(bc – b^2 – c^2 + bc + a^2 + ac + ab – bc)
= -(a + b + c)(a^2 + b^2 + c^2 – ab – bc – ac)
= -(a^3 + b^3 + c^3 – 3abc)