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Evaluation of a Determinant

Using propert...

Question

Using properties of determinants, show that
$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (b + c)^{2} & a b & c a a b & (c + a)^{2} & b c a c & b c & (a + b)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2 a b c (a + b + c)^{2}$

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∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & b c & a c + c^{2} a^{2} + a b & b^{2} & a c a b & b^{2} + b c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 4 a^{2} b^{2} c^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c c a & c b & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 1 + a^{2} + b^{2} + c^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} {- a}^{2} & a b & a c b a & {- b}^{2} & b c c a & c b & {- c}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {4 a}^{2} b^{2} c^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c c a & c b & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 1 + a^{2} + b^{2} + c^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c b a & b^{2} + 1 & b c c a & c b & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a^{2} + b^{2} + c^{2} + 1

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