The value of the determinant

$\left|\begin{array}{lll}a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c\end{array}\right|$

$=\left|\begin{array}{lll}-b & b+c+a & a \\ -c & c+a+b & b \\ -a & a+b+c & c\end{array}\right|$ $\left[\right.$ Applying $C_{1} \rightarrow C_{1}-C_{3}$ and $\left.C_{2} \rightarrow C_{2}+C_{3}\right]$

$=(-1)(a+b+c)\left|\begin{array}{lll}b & 1 & a \\ c & 1 & b \\ a & 1 & c\end{array}\right|$           [Taking $(-1)$ common from $C_{1}$ and $(a+b+c)$ common from $C_{2}$ ]

$=(-1)(a+b+c)\left|\begin{array}{ccc}b & 1 & a \\ c-b & 0 & b-a \\ a-b & 0 & c-a\end{array}\right|$          [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ ]

$=(-1)(a+b+c)[-(c-b)(c-a)+(b-a)(a-b)]$

$=(-1)(a+b+c)\left[-c^{2}+a c+b c-a b+b a-b^{2}-a^{2}+a b\right]$

$=(-1)(a+b+c)\left(-a^{2}-b^{2}-c^{2}+a b+b c+a c\right)$

$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)$

$=a^{3}+a b^{2}+a c^{2}-a^{2} b-a b c-a^{2} c+b a^{2}+b^{3}+b c^{2}-a b^{2}-b^{2} c-a b c+c a^{2}+c b^{2}+c^{3}-a c b-b c^{2}-a c^{2}$

$=a^{3}+b^{3}+c^{3}-3 a b c$

Hence, the correct option is (c).