1
$\begingroup$

What method should I use to solve this system of polynomial equations? $$\begin{cases} 3x^3 - 3y^3 + z^3 - xyz - 3 &= 0\\ 3y^3 - x^3 - z^3 - xyz + 5 &= 0\\ x^3 - y^3 + z^3 - xyz - 2 &= 0\end{cases}$$


I've run out of ideas. Tried adding, subtracting but it gave me nothing. Maybe there's a good substitution I don't see?

$\endgroup$
4
  • $\begingroup$ Is this correctly writen? $\endgroup$ – Aqua Mar 29 '20 at 8:57
  • $\begingroup$ yes, i've checked it $\endgroup$ – Joseph Larson Mar 29 '20 at 8:59
  • 1
    $\begingroup$ Is it over $\Bbb C$? Or over integers, or reals, or rationals? $\endgroup$ – Dietrich Burde Mar 29 '20 at 9:03
  • $\begingroup$ i'm not sure actually this is not specified in the problem statement. i guess reals/rationals not integers $\endgroup$ – Joseph Larson Mar 29 '20 at 9:08
1
$\begingroup$

Hint. By summing the first two equations we obtain $$xyz=x^3+1.$$ Then solve the linear system with respect to $y^3$ and $z^3$: $$\begin{cases} 3x^3 - 3y^3 + z^3 - (x^3+1) - 3 = 0\\ x^3-y^3+z^3-(x^3+1)-2=0 \end{cases}$$ and we find $$y^3=x^3-\frac{1}{2},\quad z^3=x^3+\frac{5}{2}.$$ Going back to the first equation cubed we get $$x^3\left(x^3-\frac{1}{2}\right)\left( x^3+\frac{5}{2}\right)=(x^3+1)^3.$$ Can you take it from here?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.