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

0

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a^{3} + b^{3} + c^{3} - 3 a b c

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3 a b c

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(a + b + c)^{3}

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Solution

The correct option is A $0$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} - b c 1 & b & b^{2} - a c 1 & c & c^{2} - a b \end{matrix} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & a - b & (a - b) (a + b + c) 0 & b - c & (b - c) (a + b + c) 1 & c & c^{2} - a b \end{matrix} ∣ ∣ ∣ ∣ ∣$
by ${\begin{matrix} R_{1} \to R_{1} - R_{2} R_{2} \to R_{2} - R_{3} \end{matrix} = (a - b) (b - c) ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & a + b + c 0 & 1 & a + b + c 1 & c & c^{2} - a b \end{matrix} ∣ ∣ ∣ ∣ = 0, {∵ R_{1} \equiv R_{2}}$ .

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