The roots $z_{1}$ , $z_{2}$ , $z_{3}$ of the equation $x^{3} + 3 a x^{2} + 3 b x + c = 0$ in which a, b, c are complex numbers, correspond to the points A,B, C on the complex plane. Then ABC is an equilateral triangle if

a^{2} = b

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a = b^{2}

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a + b^{2} = 0

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a^{2} + b = 0

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Solution

The correct option is B $a^{2} = b$
For $z_{1} z_{2} z_{3} t o f o r m a n e q u i l a t e r a l △$ ,
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} - z_{1} z_{2} - z_{2} z_{3} - z_{3} z_{1} = 0 . . . . . . . . . . . (1)$
Now,
$z_{1} + z_{2} + z_{3} = - 3 a [s u m o f r o o t s = \frac{- (c o e f f i c i e n t o f x^{2})}{c o e f f i c i e n t o f x^{3}}]$
$z_{1} z_{2} + z_{2} z_{3} + z_{1} z_{3} = 3 b [s u m o f p r o d u c t o f r o o t s t a k e n t w o a t a t i m e = \frac{c o e f f i c i e n t o f x}{c o e f f i c i e n t o f x^{3}}]$
$z_{1} z_{2} z_{3} = - c [p r o d u c t o f r o o t s = \frac{- (c o n s t a n t t e r m)}{c o e f f i c i e n t o f x^{3}}]$
Now,
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = {(z_{1} + z_{2} + z_{2})}_{1}^{2} - 2 (z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1})$
$= 9 a^{2} - 2 (3 b) . . . . . . . . . . . . (3)$

By 1, 2 & 3:
$9 a^{2} - 6 b - 3 b = 0$
$\Rightarrow 9 a^{2} = 9 b$
$\Rightarrow a^{2} = b$

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