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

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Solution

$∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ - ∣ ∣ ∣ ∣ \begin{matrix} b & c & a c & a & b a & b & c \end{matrix} ∣ ∣ ∣ ∣ = 0$
By finding the determinants of each
we get equal to $0$

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∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a b - c & c - a & a - b c - a & a - b & b - c \end{matrix} ∣ ∣ ∣ ∣ = 0

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

= 0

Using the property of determinants and without expanding, prove that:

|\begin{matrix} a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c \end{matrix}| = 2 |\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}|

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

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