b
(b) $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = \left| {\,\begin{array}{*{20}{c}}{a + b + c}&{a + b + c}&{a + b + c}\\b&c&a\\c&a&b\end{array}\,} \right|$,
$({R_1} \to {R_1} + {R_2} + {R_3})$
=$(a + b + c)$ $\left| {\,\begin{array}{*{20}{c}}{1}&{1}&{1}\\b&c&a\\c&a&b\end{array}\,} \right|$ =$(a + b + c)$ $\left| {\,\begin{array}{*{20}{c}} 1&1&1 \\ b&c&a \\ c&a&b \end{array}\,} \right|$ = $ (a + b + c)$ $\left| {\,\begin{array}{*{20}{c}} 1&0&0 \\ b&{b - c}&{c - a} \\ c&{c - a}&{a - b} \end{array}\,} \right|$
= $3abc - {a^3} - {b^3} - {c^3}$, (After simplification).