Question

Using the property of determinants and without expanding.
$\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|=0$

Answer 1

Let $A=\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|=0$
Applying $C_1\to C_1+C_2+C_3$, we get
$=\left|\begin{array}{ccc}a-b+b-c+c+c-a& b-c& c-a\\ b-c+c-a+a-b& c-a& a-b\\ c-a+a-b+b-c& a-b& b-c\end{array}\right|=\left|\begin{array}{ccc}0& b-c& c-a\\ 0& c-a& a-b\\ 0& a-b& b-c\end{array}\right|=0$
[Since, each element of $C_1$ (first column ) is zero].

Alternate method
$A=\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|$ Applying $R_1\to R_1+R_2,$ we get
$=\left|\begin{array}{ccc}a-c& b-a& c-b\\ b-c& c-a& a-b\\ -(a-c)& -(b-a)& -(c-b)\end{array}\right|$
Taking negative sign common from $R_3$, we get
$=-\left|\begin{array}{ccc}a-c& b-a& c-b\\ b-c& c-a& a-b\\ a-c& b-a& c-b\end{array}\right|=0$ [Since, the two rows $R_1$ and $R_3$ are identical ].