$∣ ∣ ∣ ∣ \begin{matrix} b + c & a - c & a - b b - c & c + a & b - a c - b & c - a & a + b \end{matrix} ∣ ∣ ∣ ∣ =$

4 a b c

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6 a b c

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8 a b c

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2 a b c

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Solution

The correct option is C $8 a b c$
$∣ ∣ ∣ ∣ \begin{matrix} b + c & a - c & a - b b - c & c + a & b - a c - b & c - a & a + b \end{matrix} ∣ ∣ ∣ ∣$

$R_{1} \to R_{1} + R_{2}$

$∣ ∣ ∣ ∣ \begin{matrix} 2 b & 2 a & 0 b - c & c + a & b - a c - b & c - a & a + b \end{matrix} ∣ ∣ ∣ ∣$

$R_{2} \to R_{2} + R_{3}$

$∣ ∣ ∣ ∣ \begin{matrix} 2 b & 2 a & 0 0 & 2 c & 2 b c - b & c - a & a + b \end{matrix} ∣ ∣ ∣ ∣$

$= 2 b (2 c (a + b) + 2 b (a - c)) + 2 a (2 b (c - b))$
$= 2 b (2 a c + 2 b c + 2 a b - 2 b c) + 2 a (2 b (c - b))$
$= 4 (b (a c + a b)) + 4 (a b (c - b))$
$= 4 a b c + 4 a b^{2} + 4 a b c - 4 a b^{2}$
$= 8 a b c$

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