A non linear system: x+y+z=2, x^2+y^2+z^2=14, x^4+y^4+z^4=98

[PoCTo]

Maybe some ideas, tricks?

[Honeysuckle588]

Well...graphing it showed me there are 6 points that solve the system. Can you first determine the intersection of the plane and the sphere? It's a circle, and if you can get the equation for that you can then work on finding the intersection of the circle with the cube-like 4th degree.

[stapel_eliz]

Sorry to resurrect such an old thread, but questions of this sort do arise from time to time, and this one's been bugging me for a while....

We have the following relations:

. . . . .

Then:

. . . . .

And plugging in gives us:

. . . . .

. . . . .

If one does the long multiplication, one will find that:

. . . . .

. . . . . . . . .

Then plugging in gives us:

. . . . .

. . . . .

Restating:

. . . . .

...as:

. . . . .

...and squaring, we get:

. . . . .

. . . . . . . . .

. . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . . . .

Substitution yields:

. . . . .

. . . . .

. . . . .

But this says that:

. . . . .

...which clearly is not a possible value for the sum of squares of real numbers. Now what...?

[PoCTo]

It's too late to understand where is the mistake, I can only write my own solution to this now
It's similar to yours, I think

Let:




Then:



(After this we can get new roots, so we have to check them after solving)

So we know:
, ,

We know, that these formulas are cube polynomial's coefficients.
If is the polynom:
, so are roots of this polynom

Now we can use some formulas (http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function) or find trivial solutions, they are:

Then we must understand that all variables are equivalent, have the same rights, so if we swap some of them, we will get a new answer.
Finally the answers are: