challenging question..?

Question

if a, b & c are positive real constants. Find the positive real solutions x, y & z in terms of a, b, & c in the equations:

x + y + z = a + b + c
4xyz - (a²x + b²y + c²z) = abc

Answer 1

Let x=a, y=b, z=c it satisfy the equation( i)
put this value in (ii)
4abc -(a^3 +b^3 + c^3) = abc
(a^3 +b^3 + c^3) = 3abc
which shows if a+b+c=0 then (a^3 +b^3 + c^3) = 3abc
Therefore
x= a= -(b+c)
y =b = -(c+a)
z =c = -(a+b)
where a+b+c=0
There may be other solutions also
x= b= -(a+c)
y =c = -(b+a)
z =a = -(c+b) and
x= c= -(b+a)
y =a = -(c+b)
z =b = -(a+c)
Other method let z=0
and solve
x+y=a+b+c
- (a²x + b²y) = abc
solve then
x={a(a^2 +ab+bc+ca)}/(a^2 -b^2)
y={b(b^2 +ab+bc+ca)}/(b^2 -a^2)

Answer 2

3 unknowns, but only 2 equations.

No unique solution exists.

Answer 3

Yes indeed, most challenging. I shall wait in earnest anticipation for a correct answer.

Answer 4

i dont believe you can figure it out because we have 6 exponents but no solutions to any of them besides another exponent.