If in a triangle $A B C, ∠ B = 60^{0},$ then

{(a - b)}^{2} = c^{2} - a b

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{(b - c)}^{2} = a^{2} - b c

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{(c - a)}^{2} = b^{2} - a c

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

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Solution

The correct option is D $a^{2} + b^{2} + c^{2} = 2 b^{2} + a c$
$∠ B = 60^{0}$
$cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$
$\Rightarrow cos 60^{0} = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$
$\Rightarrow \frac{1}{2} = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$
$\Rightarrow a c = a^{2} + c^{2} - b^{2}$
Add $2 b^{2}$ both sides, we get
$\Rightarrow 2 b^{2} + a c = 2 b^{2} + a^{2} + c^{2} - b^{2}$
$∴ a^{2} + b^{2} + c^{2} = 2 b^{2} + a c$

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∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} + a^{4} & 1 + a b + a^{2} b^{2} & 1 + a c + a^{2} c^{2} 1 + a b + a^{2} b^{2} & 1 + b^{2} + b^{4} & 1 + b c + b^{2} c^{2} 1 + a c + a^{2} c^{2} & 1 + b c + b^{2} c^{2} & 1 + c^{2} + c^{4} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(a - b)}^{2} {(b - c)}^{2} {(c - a)}^{2}

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