Using the property of determinants and without expanding, prove that:
$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$
Using the property of determinants and without expanding, prove that:
$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$
$\Delta=\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2},$ we have:
$\Delta=\left|\begin{array}{ccc}a-c & b-a & c-b \\ b-c & c-a & a-b \\ -(a-c) & -(b-a) & -(c-b)\end{array}\right|$
$=-\left|\begin{array}{ccc}a-c & b-a & c-b \\ b-c & c-a & a-b \\ a-c & b-a & c-b\end{array}\right|$
Here, the two rows $R_{1}$ and $R_{3}$ are identical.
$\therefore \Delta=0$
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{{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\
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