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

abc

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4abc

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4a²b²c²

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a²b²c²

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Solution

The correct option is C 4a²b²c²
$△ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} + c^{2} & a^{2} & a^{2} b^{2} & c^{2} + a^{2} & b^{2} c^{2} & c^{2} & a^{2} + b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 2 ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & c^{2} & b^{2} b^{2} & c^{2} + a^{2} & b^{2} c^{2} & c^{2} & a^{2} + b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣, b y R_{1} \to R_{1} - (R_{2} + R_{3}) = - 2 ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & c^{2} & b^{2} b^{2} & a^{2} & 0 c^{2} & 0 & a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣, b y \begin{matrix} R_{2} \to R_{2} - R_{1} R_{3} \to R_{3} - R_{1} \end{matrix} = - 2 {- c^{2} (b^{2} a^{2}) + b^{2} (- c^{2} a^{2})} = 4 a^{2} b^{2} c^{2} .$
Trick: Put a=1, b=2, c=3, so that the option give different values.

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Similar questions

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

\frac{a^{2} + a b + b^{2}}{b^{2} + b c + c^{2}} = \frac{a}{c}

\frac{a^{4} + a^{2} b^{2} + b^{4}}{b^{4} + b^{2} c^{2} + c^{4}} = \frac{a^{2}}{c^{2}}

a^{4} + b^{4} + c^{4} \geq a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}

a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} \geq a b c (a + b + c)

Δ A B C

c^{4} - 2 (a^{2} + b^{2}) c^{2} + a^{4} + a^{2} b^{2} + b^{4} = 0

∠ C =

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