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

(a + b + c)^{3}

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

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

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

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Solution

The correct option is A $(a + b + c)^{3}$
Let $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣$
Since the answer is $(a + b + c)^{3}$ , we shall try to get $(a + b + c)$ .
Applying $R_{1} \to R_{1} + R_{2} + R_{3}$ , then
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} a + b + c & a + b + c & a + b + c 2 b & b - c - a & 2 b 2 c & 2 c & c - a b \end{matrix} ∣ ∣ ∣ ∣$
Taking (a + b + c) common from $R_{1}$ , we get
$Δ = (a + b + c) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣$
Applying $C_{3} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$
$∴ Δ = (a + b + c) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 2 b & - a - b - c & 0 2 c & 0 & - c - a - b \end{matrix} ∣ ∣ ∣ ∣$
(By property since all elements above leading diagonal are zero)
$= (a + b + c) . 1. (- a - b - c) . (- c - a - b)$
Hence $Δ = (a + b + c)^{3}$

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