The roots of the equation $t^{3} + 3 a t^{2} + 3 b t + c = 0$ are $z_{1}$ , $z_{2}$ , $z_{3}$ which represent the vertices of an equilateral triangle. Then

a^{2} = 3 b

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b^{2} = a

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a^{2} = b

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b^{2} = 3 a

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Solution

The correct option is B $a^{2} = b$
Using theory of equations.
$z_{1} + z_{2} + z_{3} = - 3 a$
$z_{1} z_{2} + z_{3} z_{2} + z_{1} z_{3} = 3 b$
$z_{1} z_{2} z_{3} = - c$

Now using formula of $(a^{2} + b^{2} + c^{2})$ = $(a + b + c)^{2} - 2 (a b + a c + b c)$
$(z_{1}^{2} + z_{2}^{2} + z_{3}^{2})$ = $(z_{1} + z_{2} + z_{3})^{2} - 2 (z_{1} z_{2} + z_{3} z_{2} + z_{1} z_{3})$
$9 a^{2} - 2 (3 b)$
$9 a^{2} - 6 b$

Hence for equilateral triangle.
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{3} z_{2} + z_{1} z_{3}$
$9 a^{2} - 6 b = 3 b$
$9 a^{2} = 9 b$
$a^{2} = b$

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