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Mathematics Question on Determinants

Using the property of determinants and without expanding, prove that: ∣a−bb−cc−a\b−cc−aa−b−¸aa−bb−c∣\begin{vmatrix}a-b&b-c&c-a\\\b-c&c-a&a-b\\\c-a&a-b&b-c\end{vmatrix}=0

Answer

△\triangle= ∣a−bb−cc−a\b−cc−aa−b−¸aa−bb−c∣\begin{vmatrix}a-b&b-c&c-a\\\b-c&c-a&a-b\\\c-a&a-b&b-c\end{vmatrix}

Applying R1→\to R1+R2,we have

△\triangle= ∣a−cb−ac−b\b−cc−aa−b−(a−c)−(b−a)−(c−b)∣\begin{vmatrix}a-c&b-a&c-b\\\b-c&c-a&a-b\\\\-(a-c)&-(b-a)&-(c-b)\end{vmatrix}

= -\begin{vmatrix}a-c&b-a&c-b\\\b-c&c-a&a-b\\\(a-c)&(b-a)&(c-b)\end{vmatrix}

Here, the two rows R1 and R3 are identical.
△\triangle = 0.