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PROVE THAT THE DETERMINANT

b²+c² ab ac

ab c² +a²bc

ac bc a²+b²

is equal to 4a²b²c²

$L e t Δ = | \begin{matrix} b^{2} + c^{2} & a b & a c \\ b a & a^{2} + c^{2} & b c \\ c a & c b & a^{2} + b^{2} \end{matrix} | M u l t i p l y R_{1}, R_{2} a n d R_{3} b y a, b & c, w e g e t, Δ = \frac{1}{a b c} | \begin{matrix} a (\begin{matrix} b^{2} + c^{2} \end{matrix}) & a^{2} b & a^{2} c \\ b^{2} a & b (\begin{matrix} a^{2} + c^{2} \end{matrix}) & b^{2} c \\ c^{2} a & c^{2} b & c (\begin{matrix} a^{2} + b^{2} \end{matrix}) \end{matrix} | = \frac{a b c}{a b c} | \begin{matrix} (\begin{matrix} b^{2} + c^{2} \end{matrix}) & a^{2} & a^{2} \\ b^{2} & (\begin{matrix} a^{2} + c^{2} \end{matrix}) & b^{2} \\ c^{2} & c^{2} & (\begin{matrix} a^{2} + b^{2} \end{matrix}) \end{matrix} | [Taking a,b,c common from C_{1}, C_{2} & C_{3}] = | \begin{matrix} 2 (\begin{matrix} b^{2} + c^{2} \end{matrix}) & 2 (a^{2} + c^{2}) & 2 (a^{2} + b^{2}) \\ b^{2} & (\begin{matrix} a^{2} + c^{2} \end{matrix}) & b^{2} \\ c^{2} & c^{2} & (\begin{matrix} a^{2} + b^{2} \end{matrix}) \end{matrix} | [A p p l y i n g R_{1} \to R_{1} + R_{2} + R_{3}] = 2 | \begin{matrix} \begin{matrix} b^{2} + c^{2} \end{matrix} & a^{2} + c^{2} & a^{2} + b^{2} \\ b^{2} & \begin{matrix} a^{2} + c^{2} \end{matrix} & b^{2} \\ c^{2} & c^{2} & \begin{matrix} a^{2} + b^{2} \end{matrix} \end{matrix} |, t a k i n g 2 c o m m o n f r o m R_{1} = 2 | \begin{matrix} \begin{matrix} b^{2} + c^{2} \end{matrix} & a^{2} + c^{2} & a^{2} + b^{2} \\ {- c}^{2} & \begin{matrix} 0 \end{matrix} & {- a}^{2} \\ {- b}^{2} & {- a}^{2} & \begin{matrix} 0 \end{matrix} \end{matrix} | A p p l y i n g R_{2} \to R_{2} - R_{1} & R_{3} \to R_{3} - R_{1} = 2 | \begin{matrix} 0 & c^{2} & b^{2} \\ {- c}^{2} & \begin{matrix} 0 \end{matrix} & {- a}^{2} \\ {- b}^{2} & {- a}^{2} & \begin{matrix} 0 \end{matrix} \end{matrix} | A p p l y i n g R_{1} \to R_{1} + R_{2} + R_{3} E x p a n d i n g a l o n g R_{1}, w e g e t, = 2 [a^{2} b^{2} c^{2} + a^{2} b^{2} c^{2}] = 4 a^{2} b^{2} c^{2}$

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