$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} - b c 1 & b & b^{2} - c a 1 & c & c^{2} - a b \end{matrix} ∣ ∣ ∣ ∣ ∣$ =?

5

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a b c

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1

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0

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Solution

The correct option is C $0$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} - b c 1 & b & b^{2} - c a 1 & c & c^{2} - a b \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow R_{1} = R_{1} - R_{2}$ and $R_{2} = R_{2} - R_{3}$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & a - b & a^{2} - b^{2} + a c - b c 0 & b - c & b^{2} - c^{2} + a b - a c 1 & c & c^{2} - a b \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & a - b & (a - b) (a + b + c) 0 & b - c & (b - c) (a + b + c) 1 & c & c^{2} - a b \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= (a - b) (b - c) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & (a + b + c) 0 & 1 & (a + b + c) 1 & c & c^{2} - a b \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow R_{1} = R_{1} - R_{2}$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 0 0 & 1 & (a + b + c) 1 & c & c^{2} - a b \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= 0.$
Hence, the answer is $0.$

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