# Find the value of the determinant $\begin{vmatrix}a-b & b+c & a\\b-c & c+a & b\\c-a & a+b & c\end{vmatrix}$

$\begin{array}{1 1} \quad a^3+b^3+c^3 \\ \quad 3bc \\ \quad a^3+b^3+c^3-3abc \\ \text{none of these} \end{array}$

Toolbox:
• If A is a square matrix such that element of a row (or a column) of A is expressed as the sum of two or more terms,then the determinant of A can be expressed as the sum of the determinants of two or more matrices.
• If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
• $|A|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
Let $\Delta=\begin{vmatrix}a-b & b+c & a\\b-c & c+a & b\\c-a & a+b & c\end{vmatrix}$
Apply $R_1\rightarrow R_3-R_1$
$\Delta=\begin{vmatrix}b & b+c & a\\c & c+a & b\\a & a+b & c\end{vmatrix}$
This can be split and written as
$\Delta=\begin{vmatrix}b & b & a\\c & c & b\\a & a & c\end{vmatrix}+\begin{vmatrix}b & c & a\\c &a & b\\a & b & c\end{vmatrix}$
Hence $\Delta=\Delta_1+\Delta_2$
In $\Delta_1$ since two columns are identical its value is 0.
Therefore $\Delta=\begin{vmatrix}b & c & a\\c & a & b\\a & b & c\end{vmatrix}$
Expanding along $R_1$ we get,
$\Delta=b(ac-b^2)c(c^2-ab)+a(bc-a^2)$
$\quad=abc-b^3-c^3+abc+abc-a^3$
$\quad=3abc-(a^3+b^3+c^3)$ or $a^3+b^3+c^3-3abc$
Hence the correct answer is C.