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How can I solve this cubic equation?

$H^{3} − 3\left[(1 + A\cos(T) )^{2} + \frac{2r \cdot A \sin(T)}{B}\right]H + 2(1 + A \cos(T))^{3} = 0$

Solution in terms of H.

Edited in order to give more insight to my problem: It was an equation which comes as a part of a derivation in Computational Fluid Dynamics. My motivation is to get H in terms of A, B, r and T. And plot a graph between H and r keeping A and B and T as constants.

Thanks!

A general doubt: If a cubic equation consists of a imaginary root, then is it compulsory that the number of imaginary roots should always be 2?

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    @bala, the fundamental theorem of algebra states that a cubic equation has 3 roots. Any imaginary root must be accompanied by its complex conjugate, so a cubic either has 3 real roots or 1 real root and two complex roots.2011-03-08

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  1. Write it as $H^3 + qH + p = 0$;
  2. use Wolfram Alpha or formulas for roots;
  3. plug in for q and p.
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    @bala maverick: Yes. In general, for a polynomial with real coefficients, any complex roots come in conjugate pairs, so if $x + iy$ is a root for real $x$ and $y$ then $x - iy$ is a root. And since a cubic only has three roots (including multiple roots) there is either a conjugate pair of roots with imaginary parts or no roots with imaginary parts.2011-03-08