If in a triangle ABC, $∠ B = 60^{o}$ , 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 options are
C $(c - a)^{2} = b^{2} - a c$
D $a^{2} + b^{2} + c^{2} = 2 b^{2} + a c$
$c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}$

$\frac{1}{2} = \frac{c^{2} + a^{2} - b^{2}}{2 c a}$

$\Rightarrow c^{2} + a^{2} = b^{2} + c a$

$\Rightarrow c^{2} + a^{2} - 2 a c = b^{2} + a c - 2 a c$

$\Rightarrow (c - a)^{2} = b^{2} - a c$

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

Hence options $[C], [D]$ are the correct answers

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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}

(- b c, b^{2} + b c, c^{2} + b c), (a^{2} + a c, - a c, c^{2} + a c)

(a^{2} + a b, b^{2} + a b, - a b)

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