# If $a+b+c\neq 0\;and\;\begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix}=0,then\;prove\;that\;a=b=c.$

Toolbox:
• If some or all elements of a row or a column of a determinant are expressed as sum of two(or more) terms,then the determinant can be expressed as sum of two (or more) determinants.
• If each element of a row (or a column)of a determinant is multiplied by a constant k,then its value gets multiplied by k.
Let $\Delta=\begin{vmatrix}a & b& c\\b & c & a\\c & a & b\end{vmatrix}$
Apply $R_1\rightarrow R_1+R_2+R_3$
$\Delta=\begin{vmatrix}a+b+c & a+b+c& a+b+c\\b & c & a\\c & a & b\end{vmatrix}$
Take (a+b+c) as the common factor from $R_1$,
$\Delta=(a+b+c)\begin{vmatrix}1 & 1& 1\\b & c & a\\c & a & b\end{vmatrix}$
Apply $C_2\rightarrow C_2-C_3$ and $C_3\rightarrow C_3-C_1$
$\Delta=(a+b+c)\begin{vmatrix}1 & 0& 0\\b & c-a & a-b\\c & a-b & b-c\end{vmatrix}$
Expanding along $R_1$ we get,
$\Delta=(a+b+c)[(c-a)(b-c)-(a-b)(a-b)]$
$\;\;=(a+b+c)(cb-c^2-ab+ac-a^2+2ab-b^2)$
$\;\;=(a+b+c)(ab+bc+ca-a^2-b^2-c^2)$
Given $\mid \Delta\mid=0$ and $(a+b+c)\neq 0$
(i.e)$(a+b+c)(ab+bc+ca-a^2-b^2-c^2)=0$
Since (a+b+c)$\neq 0$
$\Rightarrow (ab+bc+ca-a^2+b^2-c^2)=0.$
Multiply by 2
We get
$2a^2+2b^2+2c^2-2ab-2bc-2ca=0.$
We know $(a-b)^2=a^2+b^2-2ab.$
$\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0.$
[$(a-b)^2,(b-c)^2,(c-a)^2$] are non negative
Therefore (a-b)=(b-c)=(c-a)
$\Rightarrow a=b=c$
Hence proved.