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I want to solve the system of equations , $$\left\{ \begin{array}{rcl} y^3 − 9x^2 + 27x − 27 &=& 0\\ z^3 − 9y^2 + 27y − 27 &=& 0\\ x^3 − 9z^2 + 27z − 27 &=& 0\\ \end{array} \right.$$

Since I've never solved a non linear system of equations in more than 2 variables and containing different degree terms in each equation ( Here , each equation contains degree 3 , degree 2 and degree 1 terms ) , I don't know where to begin. I'm only concerned about real roots .

Actually I checked wolfram alpha and it seems that all real roots are integers , the solution is $(3,3,3)$ . However I want to know how to solve this system ( by hand )

Please help.

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  • $\begingroup$ Integer solutions, rational solutions, real, or complex solutions? $\endgroup$ – Daniel Fischer Sep 12 '13 at 12:15
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    $\begingroup$ They resemble the expansion of $(x-3)^3$ $\endgroup$ – Empy2 Sep 12 '13 at 12:19
  • $\begingroup$ @DanielFischer only real roots $\endgroup$ – A Googler Sep 12 '13 at 12:25
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    $\begingroup$ $(x,y,z)=(3,3,3)$. $\endgroup$ – Dietrich Burde Sep 12 '13 at 12:25
  • $\begingroup$ Ah, pity. Integer solutions would have made it easy. $\endgroup$ – Daniel Fischer Sep 12 '13 at 12:25
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For integer solutions:

You can rewrite the equations using

$$y^3 - 9x^2 +27x - 27 = y^3 - x^3 + (x-3)^3$$

and similarly for the two other equations. Adding the three equations then yields

$$(x-3)^3 + (y-3)^3 + (z-3)^3 = 0.$$

It is long known that in all integer solutions of the equation $a^3 + b^3 = c^3$, one of the integers must be $0$. Our equation is (modulo rearranging) exactly that, so we must have $x = 3$ or $y = 3$ or $z = 3$.

Suppose $x=3$. Then the first equation reduces to $y^3 - x^3 = 0$, hence $y^3 = 27$, hence $y = 3$. In the same way, it follows that $z = 3$. Starting from $y = 3$ or $z = 3$ yields the same. So $(3,3,3)$ is the only integer solution to the system.

Now, for any real solution, we still have by adding the equations

$$(x-3)^3 + (y-3)^3 + (z-3)^3 = 0,$$

so if one of $x,y,z$ were different from $3$, one of the three at least would be larger than $3$. Without loss of generality, let $x = 3 +\delta > 3$. Then, from the first equation, we obtain

$$\begin{align} y^3 &= x^3 - (x-3)^3\\ &= (3+\delta)^3 - \delta^3\\ &= 3^3 + 27\delta + 9\delta^2\\ &> 3^3, \end{align}$$

so also $y > 3$. The same reasoning for the second equation then yields $z > 3$, and so

$$(x-3)^3 + (y-3)^3 + (z-3)^3 > 0$$

contradicting the assumption that $(x,y,z)$ is a solution of the system.

Hence $(3,3,3)$ is the only real solution of the system.

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  • $\begingroup$ Can you show that, if $x>3$, then $y>3$? $\endgroup$ – Empy2 Sep 12 '13 at 12:53
  • $\begingroup$ Let $x = 3+\delta > 3$. Then $y^3 = x^3 - (x-3)^3 = (3+\delta)^3 - \delta^3 = 3^3 + 27\delta + 9\delta^2 > 3^3$, hence $y > 3$. Then by the same reasoning $z > 3$, then $(x-3)^3+(y-3)^3 +(z-3)^3 > 0$. Is that what you mean? $\endgroup$ – Daniel Fischer Sep 12 '13 at 13:00
  • $\begingroup$ Yay, well that's that done then. $\endgroup$ – Empy2 Sep 12 '13 at 13:02

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