x+y+z=6
x^2+y^2+z^2=8
x^3+y^3+z^3=5
x^4+y^4+z^4=?
how to find this help me soon
Based on the fact that all three of your equations are equal to whole numbers, I would GUESS that all three numbers are whole numbers. If they aren't, I find it hard to believe that summing them, summing their squares, and summing their cubes all result in whole numbers.
Also, based on the fact that when you sum their cubes you get a smaller answer than when you sum their squares, one of them (at least) must be negative - the negative sign will go away when you have an even power (like 2) but will be present when you have an odd power (like 3), thus making the sum of odd powers less than the sum of even powers.
However, assuming that all that is correct, I tried all combinations of x, y, and z, with x ranging from -20 to 20 and y ranging from -20 to 20 (and z dependent on x and y, based on the first equation) and got no solutions.
Not sure if any of this is any help. Maybe it means they answers are outside of that range, which seems very doubtful (the lowest values I got for the sums of squares and cubes where when, obviously, x and y were close to 0). Maybe it means that the answers aren't whole numbers. Maybe it means there is no solution, which is kind of the direction I'm leaning, but I have no proof.
Okay, I think there's no solution. Here's what I did, using just the first two equations you provided (I didn't deal with the cubes, but I don't think it's necessary).
Note that $\displaystyle (x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz$.
Since $\displaystyle (x + y + z)^2 = 6^2 = 36$, we know the above sum also equals 36. Also, we know the first three terms must sum to 8 (since it's just the second equation - the sum of the squares). Thus, the last three terms must sum to 28. So:
$\displaystyle 2xy + 2xz + 2yz = 28$ which gives $\displaystyle xy + xz + yz = 14$.
From $\displaystyle xy + xz + yz = 14$, factor an x out of the first time AND bring the 14 to the other side, giving:
$\displaystyle x (y+z) + yz - 14 = 0$.
Manipulating equation 1 a bit, we find that y = 6 - x - z. Let's substitute that in for y in the equation above. We get:
$\displaystyle x (6 - x - z + z) + (6 - x - z)z - 14 = 0$ which gives:
$\displaystyle 6x - x^2 + 6z - xz - z^2 - 14 = 0$.
For ease, we'll multiply everything by -1 and rearrange a bit, ordering the terms based on the power of x in them.
$\displaystyle x^2 + xz - 6x + z^2 -6z + 14 = 0$.
Now, we'll treat the x as the "variable" (even though they're all variable, really).
Rewriting this as a quadratic equation, as a function of x:
$\displaystyle x^2 + (z-6)x + (z^2 - 6z + 14) = 0$.
Almost done. This is a quadratic equation, so let's check the discriminant ($\displaystyle b^2 - 4ac$) to see how many solutions it has. Recall if the discriminant is negative, we have no real solutions.
In our equation, $\displaystyle a = 1, b = z-6, c = z^2 - 6z + 14$.
Thus, the discriminant is:
$\displaystyle (z-6)^2 - 4(1)(z^2 - 6z + 14)$.
Simplifying:
$\displaystyle z^2 - 12z + 36 - 4z^2 + 24z - 56$
$\displaystyle = -3z^2 + 12z - 18$.
Now, we want this discriminant to be positive (or equal to zero) so we have a solution. Use the inequality:
$\displaystyle -3z^2 + 12z - 18 > 0$. Divide both sides by -3 (and reverse the inequality sign):
$\displaystyle z^2 - 4z + 6 < 0$.
This is a quadratic equation with, obviously, a parabolic graph. It opens upward. The axis of symmetry, given by $\displaystyle x = \frac{-b}{2a}$ is the line x = 2. The y-coordinate can be found by plugging 2 in for z (which plays the role of x in our quadratic). This gives a y-coordinate of 2. So our parabola has a vertex of (2,2) and opens upward and is, thus, never less than or equal to 0.
So, our inequality has no solutions.
So, the discriminant is never greater than or equal to zero.
So, our original quadratic, derived from our original two equations, has no solutions.
So, our system has no solutions.
If someone can verify this is correct, that would be nice.
Ok, I just ran it coast to coast on the attached file I created from -100 to 100 with all 6 possible combinations of square roots of the integers entered as well. I used a 0.1 tolerance and I got no Solution.
IF somebody is bored tonight, they can run the attached file with limits of -500 to 500. I'm starting to lean the "No Solution" route as well.
Here is another way.
1)$\displaystyle x+y+z=6\implies (x+y+z)^2 = 36 \implies (x^2+y^2+z^2)+2(xy+xz+yz)=36$ thus $\displaystyle xy+xz+yz = 14$.
2)$\displaystyle x^2+y^2+z^2 = 8\implies (x^2+y^2+z^2)^2 = 64\implies (x^4+y^4+z^4)+2(x^2y^2+x^2z^2+y^2z^2) = 64$
3)$\displaystyle xy+xz+yz = 19\implies $$\displaystyle (xy+xz+yz)^2 = 361\implies (x^2y^2+x^2z^2+y^2z^2) + 2xyz(x+y+z) = 361$.
4)$\displaystyle x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - xz - yz) \implies xyz = \frac{31}{5}$
This gives you all the infromation to find $\displaystyle x^4+y^4+z^4$ because in #2 we need to know what $\displaystyle x^2+y^2+x^2z^2+y^2z^2$ is by in #3 we have an equation for it in terms of $\displaystyle xyz$ which we already found in #4.