Could anyone help me solve:

x+y+z=a
xy+yz+zx=b
xyz=c

Thanks.

Originally posted by goofy headed punk
Could anyone help me solve:

x+y+z=a
xy+yz+zx=b
xyz=c

Thanks.

i haven t finished the problem yet, i don t know if i will be able to get to it tonight, but i did look at it briefly, its kind of a neat problem

the first equation has a locus that is a plain that intersects each axis at a.

the second equation, together with the first, imply that the solution must also lie on a sphere of radius b²-2a.

if this sphere doesn t intersect the plane above, then there will be no (real) solution

also note that if any (h,k,l) is a solution, then so is any permutation, like (k,l,h) etc.. it is symmetric. so finding one solution will give us more solutions.

my next step would be to change coordinates to the plane, and solve the third equation in those coordinates.

This is the closest I've gotten.
x+y+z=a
xy+yz+zx=b
xyz=c

x=a-y-z, equ#1
x=c/zy, equ#3
a-y-z=c/zy, ay-yy-zy=c/z, -yy-y(z-a)=c/z, y^2+y(z-a)+c/z
solve for y:

x=a-y-z
y=1/2/z*(-z^2+z*a+&-sqrt(z^4-2*z^3*a+z^2*a^2-4*z*c))
z=t

x=-t*(t*a-&+sqrt(t^4-2*t^3*a+t^2*a^2-4*t*c)-2*b-t^2)/(t^2+t*a-&+sqrt(t^4-2*t^3*a+t^2*a^2-4*t*c)), equ#2

z=1/6*(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)-6*(1/3*b-1/9*a^2)/(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)+1/3*a, -1/12*(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)+3*(1/3*b-1/9*a^2)/(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)+1/3*a+1/2*I*sqrt(3)*(1/6*(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)+6*(1/3*b-1/9*a^2)/(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)), -1/12*(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)+3*(1/3*b-1/9*a^2)/(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)+1/3*a-1/2*I*sqrt(3)*(1/6*(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)+6*(1/3*b-1/9*a^2)/(-36*b*a+108*c+8*a^3+12*sqrt(12*b^3-3*b^2*a^2-54*b*a*c+81*c^2+12*c*a^3))^(1/3)), equ#2

I can't get nothing for y.
Is this even close?

I don't know if this helps at all, but I've come up with a few equations that seem to narrow things down a bit.

x^3-ax^2+bx-c=0
z^3-az^2+bz-c=0
y^3-ay^2+by-c=0

and

(x^3+y^3+z^3)/xyz=3

If that helps anyone at all, great.
I just want to see how all this plays out now.
Something looks really familiar about all of this, I just can't quite put my finger on it.

Wow, the equations look so simple, but when I sat down and tried to solve them, I spent an hour and got nothing (except more complicated equations).

I played around with thinking of the second equation as the expectation value of a 120deg rotation about the 1,1,1 axis. It didn't seem to get me anywhere.

Ok the mystery is solved. I set z = t and solve for y and x in terms of t and the constants.

click me (http://members.rogers.com/force1/solvedEquation.jpg) if you can't see pic.

http://members.rogers.com/force1/solvedEquation.jpg

Yes, but how does one get all that? Getting all that to pop out manually is the trick.

From theory of equations, (x, y, z) are roots of the following cubic equation.x³ - ax² + bx - c = 0The analytic solution of a cubic is a painful task, but a numerical solution is easy (try the Newton-Raphson method).

I know, rather unfortunately I'm looking for the analytical solution.

I myself will tend to deal with his by changing x, y, z into equations with x, y, z, a, b, c, and I think that will instead kill me....:D

I don't know what I'm doing here but I have been working with truth tables so, make a sort of truth table for x,y and z with a,b and c as results.
x y z a b c
0 0 0 0 0 0
0 0 1 1 0 0
0 1 0 1 0 0
0 1 1 2 1 0
1 0 0 1 0 0
1 0 1 2 1 0
1 1 0 2 1 0
1 1 1 3 3 1

x=y=z = 1. a = 3 = b. c = 1
I think that is the only solution, but I don't know why.

Like some others my solution is as follows:

x + y + z =a multiply both sides by x and rearrange gives

ax - x^2 = yx + zx

sub yx + zx = ax -x^2 into equ 2

gives yz+ax - x^2 = b

from equ 3 yz =c/x sub into above gives

c/x + ax -x^2 = b multiply both sides by x and rearrange gives:

x^3 -ax^2 +bx - c = 0

The analytical solution to the roots is beyond my ambition at the moment.

Cheers,

editted for +/- signs

There is no way I would go through the work of an analytical solution to a cubic when the successive approximations method is so easy.

If you want to do all the work, the following is the general approach. Givenx³ - ax² + bx - c = 0

a = x + y + z
b = xy + xy + yz
c = xyz

If the above are true, then x, y, z are roots of the cubic.The above can be solved by first substituting x = u + a/3 for x.

If you work out the algebra, you get u³ + pu + q = 0, where p & q are expressions in a, b, c. Given a, b, c you can work out the values for p, q.

Now substitute v = u - p/3u — Do a search for Cardano or Tartaglia, who worked this out a long time ago. I think this substitution is correct, but am not 100% certain. If it is wrong, it is close and you can probably figure out the correct substitution.

The above substitution and some algebraic manipulation results a quadratic in v³, with coefficients which are expressions in p & q.

Use the ordinary formula for roots of a quadratic to get values for v³. Find the three complex cube roots of each root of the quadratic to get values for v.

Then reverse the substitutions to get the three roots of the original cubic. Since you have 6 values for v, some of the roots might be spurious or might be equal to each other.

I recommend the successive iteration method.

I came up with this lasr night as I was thinking about something else entirely. I think it's valid.


From Mephura:

x^3-ax^2+bx-c=0
z^3-az^2+bz-c=0
y^3-ay^2+by-c=0

you can see the derivation in my last post.

from the fact that a, b, c are always the same you can infer from these equations that x=y=z

Subbing bacj into original equations gives

3x=a equ 1

3x^2=b equ 2

x^3=c equ 3

Now given any single value of any of x, y, z, a, b, c you can establish values for all the others.

ie x=2 gives a= 6, b=12, c=8 y=2 z=2

a=-3 gives x=-1=y=z b=3, c=-1

and so on....

Cheers!

Actually loojing over my solution, I see that x,y,z, cannot be <0, this will invalidate the cubic relationships from above. I believe this is the only limitation, though.