In this special case, if you let A = ((x^2 + 10*y^2)/10^(1/3) (assuming I did
the algebra right, note there are 3 such A's, but only 1 real one) then you get
sage: A = var("A"); solve(2* x^2 + y^2 + z^2 - 1- A*z==0,z)
[z == (A - sqrt(A^2 - 4*y^2 - 8*x^2 + 4))/2, z == (sqrt(A^2 - 4*y^2 -
8*x^2 + 4) + A)/2]
On Sat, Jun 21, 2008 at 9:43 AM, Maximilian Lepik
<[EMAIL PROTECTED]> wrote:
> ok thanks. Do you no any other way how I could solve it
>
> 2008/6/21 David Joyner <[EMAIL PROTECTED]>:
>>
>> This is a 6th degree polynomial in z. There is no general formula for
>> algebraically
>> solving for roots of polynomials of degree 5 or higher.
>>
>>
>> On Fri, Jun 20, 2008 at 8:20 PM, Max <[EMAIL PROTECTED]>
>> wrote:
>> >
>> > Hi,
>> > i am looking for a way to solve this equation "10* (2* x^2 + y^2 + z^2
>> > - 1)^3- x^2*z^3 - 10*y^2* z^3=0" to "z" but I did not find a way to do
>> > it in sage. So if anybody knows how to do it please answer me- Thanks.
>> >
>> > >
>> >
>>
>>
>
>
>
> --
> Maximilian Lepik
> 235 E. 105th Street Apartment 5C
> 10029 New York, NY
> +1-212 722 79 67 (Home US)
> +1 917 428 6238 (Mobile US)
> >
>
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