Add the product
$a (a - b), b (b - c), c (c - a)$

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Solution

Adding the product gives $a (a - b) + b (b - c) + c (c - a) = a^{2} + b^{2} + c^{2} - a b - b c - c a = \frac{1}{2} (2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a)$
$= \frac{1}{2} (a^{2} + b^{2} - 2 a b + b^{2} + c^{2} - 2 b c + c^{2} + a^{2} - 2 c a) = \frac{1}{2} ((a - b)^{2} + (b - c)^{2} + (c - a)^{2})$

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a (a - b), b (b - c), c (c - a)

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

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} ∣ ∣ ∣ ∣ = 0

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

D^{'} = ∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b a + b & b + c & c + a c + a & a + b & b + c \end{matrix} ∣ ∣ ∣ ∣

D^{'} =

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