Re: cubic equation solver

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• Subject: [mg130250] Re: cubic equation solver
• From: "Kevin J. McCann" <kjm at KevinMcCann.com>
• Date: Fri, 29 Mar 2013 05:55:31 -0400 (EDT)
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These ARE the exact roots, although not in a very convenient form. If
you plug them back in to the LHS, you get {0,0,0}.

Kevin

On 3/28/2013 4:05 AM, Elim Qiu wrote:
> x^3 + (=E2=88=9A6 + 2=E2=88=9A3 + 2=E2=88=9A2 -9)x + 2=E2=88=9A3 -=E2=88=9A2 -2 = 0
> has exact roots =E2=88=9A2-2, =E2=88=9A3-=E2=88=9A2, 2-=E2=88=9A3
>
> But Mathematica says:
>
> Solve[x^3 + (Sqrt[6] + 2 Sqrt[3] + 2 Sqrt[2] - 9) x + 2 Sqrt[3] -
>     Sqrt[2] - 2 == 0, x]
>
> {{x -> (1/
>        2 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>          I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>              264 Sqrt[6])]))^(1/3)/3^(
>      2/3) - (-9 + 2 Sqrt[2] + 2 Sqrt[3] + Sqrt[
>       6])/(3/2 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>          I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>              264 Sqrt[6])]))^(
>      1/3)}, {x -> -(((1 + I Sqrt[3]) (1/
>          2 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>            I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>                264 Sqrt[6])]))^(1/3))/(
>       2 3^(2/3))) + ((1 - I Sqrt[3]) (-9 + 2 Sqrt[2] + 2 Sqrt[3] +
>         Sqrt[6]))/(
>      2^(2/3) (3 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>           I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>               264 Sqrt[6])]))^(
>       1/3))}, {x -> -(((1 - I Sqrt[3]) (1/
>          2 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>            I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>                264 Sqrt[6])]))^(1/3))/(
>       2 3^(2/3))) + ((1 + I Sqrt[3]) (-9 + 2 Sqrt[2] + 2 Sqrt[3] +
>         Sqrt[6]))/(
>      2^(2/3) (3 (18 + 9 Sqrt[2] - 18 Sqrt[3] +
>           I Sqrt[3 (4662 - 1252 Sqrt[2] - 1296 Sqrt[3] -
>               264 Sqrt[6])]))^(1/3))}}
>

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