Using the properties of determinant and without expanding , prove that:
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a (b + c) 1 & c a & b (c + a) 1 & a b & c (a + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

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Solution

$∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a b 1 & c a & b c 1 & a b & c a \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a c 1 & c a & b a 1 & a b & c b \end{matrix} ∣ ∣ ∣ ∣ = 0$
$C_{2} \leftrightarrow C_{3}$ of first Determinant
$∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a b 1 & a b & c a 1 & c a & b c \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a c 1 & c a & b a 1 & a b & c b \end{matrix} ∣ ∣ ∣ ∣ = 0$
$C_{1} \leftrightarrow C_{2}$ of first Determinant
$∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & c a 1 & b c & a b 1 & c a & b c \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a c 1 & c a & b a 1 & a b & c b \end{matrix} ∣ ∣ ∣ ∣ = 0$
By finding determinant of this both and on cancelling we get zero

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