Re: cubic complex

• To: mathgroup at smc.vnet.net
• Subject: [mg30748] Re: [mg30728] cubic complex
• From: BobHanlon at aol.com
• Date: Sun, 9 Sep 2001 03:26:43 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```In a message dated 2001/9/8 3:12:28 AM, dmerdol at mech.ubc.ca writes:

>I am after the analytical solution for
>      a0 * x^3 + a1 * x^2 + a2 * x + a3 = 0 .
>(ai), because they complex numbers.
>

Solve[a0 * x^3 + a1 * x^2 + a2 * x + a3 == 0, x]//Simplify

{{x -> (1/(6*a0))*(-2*a1 + 2^(2/3)*
(-2*a1^3 + 9*a0*a2*a1 - 27*a0^2*a3 +
Sqrt[(2*a1^3 - 9*a0*a2*a1 + 27*a0^2*a3)^2 -
4*(a1^2 - 3*a0*a2)^3])^(1/3) +
(2*2^(1/3)*(a1^2 - 3*a0*a2))/
(-2*a1^3 + 9*a0*a2*a1 - 27*a0^2*a3 +
Sqrt[(2*a1^3 - 9*a0*a2*a1 + 27*a0^2*a3)^2 -
4*(a1^2 - 3*a0*a2)^3])^(1/3))},
{x -> (1/(12*a0))*(-4*a1 + I*2^(2/3)*(I + Sqrt[3])*
(-2*a1^3 + 9*a0*a2*a1 - 27*a0^2*a3 +
Sqrt[(2*a1^3 - 9*a0*a2*a1 + 27*a0^2*a3)^2 -
4*(a1^2 - 3*a0*a2)^3])^(1/3) -
(2*I*2^(1/3)*(-I + Sqrt[3])*(a1^2 - 3*a0*a2))/
(-2*a1^3 + 9*a0*a2*a1 - 27*a0^2*a3 +
Sqrt[(2*a1^3 - 9*a0*a2*a1 + 27*a0^2*a3)^2 -
4*(a1^2 - 3*a0*a2)^3])^(1/3))},
{x -> -((1/(12*a0))*(4*a1 + 2^(2/3)*(1 + I*Sqrt[3])*
(-2*a1^3 + 9*a0*a2*a1 - 27*a0^2*a3 +
Sqrt[(2*a1^3 - 9*a0*a2*a1 + 27*a0^2*a3)^2 -
4*(a1^2 - 3*a0*a2)^3])^(1/3) +
(2*2^(1/3)*(1 - I*Sqrt[3])*(a1^2 - 3*a0*a2))/
(-2*a1^3 + 9*a0*a2*a1 - 27*a0^2*a3 +
Sqrt[(2*a1^3 - 9*a0*a2*a1 + 27*a0^2*a3)^2 -
4*(a1^2 - 3*a0*a2)^3])^(1/3)))}}

Bob Hanlon
Chantilly, VA  USA

```

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