Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$
$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$
(N/A) Let $\Delta = \left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$.
Applying the operation $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$:
$\Delta = \left|\begin{array}{ccc}(a-b) + (b-c) + (c-a) & (b-c) + (c-a) + (a-b) & (c-a) + (a-b) + (b-c) \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$
Simplifying the elements of the first row:
$\Delta = \left|\begin{array}{ccc}0 & 0 & 0 \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$
Since all elements of the first row $R_{1}$ are $0$,the value of the determinant is $0$.
$\therefore \Delta = 0$.
Applying the operation $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$:
$\Delta = \left|\begin{array}{ccc}(a-b) + (b-c) + (c-a) & (b-c) + (c-a) + (a-b) & (c-a) + (a-b) + (b-c) \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$
Simplifying the elements of the first row:
$\Delta = \left|\begin{array}{ccc}0 & 0 & 0 \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$
Since all elements of the first row $R_{1}$ are $0$,the value of the determinant is $0$.
$\therefore \Delta = 0$.
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