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This question already has an answer here:

Problem :

The sum of non real roots of the polynomial equation $x^3+3x^2+3x+3=0$

(a) equals 0

(b) lies between 0 and 1

(c)lies between -1 and 0

(d) has absolute value bigger than 1

My approach :

The discriminant of cubic equation $ax^3+bx^2+cx+d=0$ is given by

$\Delta = 18abcd -4b^3d +b^2c^2 -4ac^3 -27a^2d^2$

$\Delta = -108 < 0 $ Therefore the equation has one real and two non real roots. But how to find the roots of this equation not getting any idea please help . thanks

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marked as duplicate by tatan, Scientifica, Shailesh, José Carlos Santos, Parcly Taxel Oct 30 '18 at 12:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Since $$(x+1)^3=-2,$$ we know one real solution is $$x=-1-\sqrt[3]{2}.$$Hence done.

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  • $\begingroup$ I deleted my answer anyways. $\endgroup$ – Shailesh Oct 30 '15 at 3:26
  • $\begingroup$ I don't think it was necessary to delete your more detailed(than mine)answer.Regards, $\endgroup$ – Arpit Kansal Oct 30 '15 at 3:48
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Hint: The equation can be written as $$(x+1)^3=-2.$$ Hence the three roots are $\sqrt[3]2-1$, $\sqrt[3]2\omega-1$, $-\sqrt[3]2\omega-1$, where $\omega$ is a cubic root of unit.

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