Multiply the first equation x^2+xy+y^2=39 by x-y, using the difference of cubes identity x^3-y^3 = (x-y)(x^2+xy+y^2), then the first equation is transformed

Similarly, multiply the second equation by y-z, then we get

Multiply the third equation by z-x

Addition of the three equations gives

39(x-y)+49(y-z)+19(z-x)=0

Expand the brackets and simplify

2x+y-3z=0, then

z^2+z

or

Substitute (4) to the first two equations x^2+xy+y^2=39, y^2+yz+z^2=49

x^2+x(3z-2x)+(3z-2x)^2=39

Simplifying gives the equation

x^2-3zx+3z^2 = 13

Now we have the system of equations

To solve the system of equations, multiply the first equation by 19 and the second one by 13. Then the constant is eliminated and we get the following equation.

19(x^2-3zx+3z^2) = 13(z^2+zx+x^2)

Simplifying the equation gives

6x^2-70zx+44z^2=0

Cancel the common factor

Factor the expression

(3x-2z)(x-11z)=0

Then we get

or

Substitute x=\dfrac{2}{3}z to the third equation z^2+zx+x^2=19, then

z^2+\dfrac{2}{3}z^2+\dfrac{4}{9}z^2 = 19

19z^2 = 19\cdot 9

Substitute to (7), we get

Substitute (9) and (10) to (4), we get

So we get two solution set \{2,3,5\} and \{-2,-3,-5\}

Substitute x=11z to the third equation z^2+zx+x^2=19, then

z^2+11z^2+121z^2 = 19

133z^2 = 19

Cancel the common factor 19, we get

z^2 = \dfrac{1}{7}

Solve for z

Substite to (8) x = 11z , then we get

Substitute both (12) and (13) to (4), we get

y = 3z-2x

= 3\Big( \pm\dfrac{\sqrt{7} }{7}\Big) -2\Big( \pm\dfrac{11\sqrt{7} }{7} \Big)

=\pm\dfrac{19\sqrt{7} }{7}

Therefore, we get another two solution set \{ \dfrac{11\sqrt{7} }{7}, -\dfrac{19\sqrt{7} }{7}, \dfrac{\sqrt{7} }{7} \} and \{ -\dfrac{11\sqrt{7} }{7}, \dfrac{19\sqrt{7} }{7}, -\dfrac{\sqrt{7} }{7} \}

Overall, there are 4 solution sets for the system of equations