Solve system of 3 equations

Question

$x+y+z=0$

$x^2+y^2+z^2=6ab$

$x^3+y^3+z^3=3(a^3+b^3)$

this is what i reasoned out so far;

$xyz=a^3+b^3$

$x^2+zx+z^2=3ab$

$y^2+zy+z^2=3ab$

$x^2+xy+y^2=3ab$

$y^2=3ab+zx$

$x^2=3ab+zy$

$z^2=3ab+xy$

I'd prefer a hint rather than a full answer - and how should I solve this kind of systems?

Answer 1

A mechanical technique is offered by "Myself" in the comments. I'll sketch it.

Via Newton's identities we have

$$\begin{array}{ll} x^2+y^2+z^2 & =(x+y+z)^2-2(xy+yz+zx) \\ x^3+y^3+z^3 & =(x+y+z)^3-3(xy+yz+zx)(x+y+z)+3xyz \end{array}$$

(Read the background for information on how these can be obtained.)

Combining with the givens and Vieta's formulas, we have

$$(t-x)(t-y)(t-z)=t^3-3abt-(a^3+b^3).$$

Thus, $x,y,z$ are the roots of this polynomial. If you want to go further and obtain explicit formulas, you can use Cardano's method for the depressed cubic.

Answer 2

If we set $z=-x-y$, then we obtain two equations in $x,y$, namely \begin{align*} 3ab - x^2 - xy - y^2 & = 0,\\ a^3 + b^3 + x^2y + xy^2 & = 0. \end{align*} For $b\neq 0$ the first equation here gives $a=\frac{x^2+xy+y^2}{3b}$, so that the second equation becomes $$ (3b^2 + 3bx + 3by + x^2 + xy + y^2)(3b^2 - 3bx + x^2 + xy + y^2)(3 b^2 - 3by + x^2 + xy + y^2)=0. $$ Each factor is a quadratic equation, say , in $y$, which can be solved easily. Since there should not be a full answer, I suggest for you to consider the case $b=0$ (for $a=b=0$ we obtain $x=y=z=0$).