Question
If x^2+xy+y^2=39, y^2+yz+z^2=49, z^2+zx+x^2=19, find all soltion sets for of x, y , z
If x^2+xy+y^2=39, y^2+yz+z^2=49, z^2+zx+x^2=19, find all soltion sets for of x, y , z
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