For real numbers $a,b,c$ solve the following system of equations:

\begin{split} a(b^2 + c) = c(c+ab) \end{split} \begin{split} b(c^2 +a) = a(a+bc) \end{split} \begin{split} c(a^2 + b) = b(b+ac) \end{split}

There are many possibilities to solve this system of equations, but the recommended solutions include algebraic manipulations which may not come to my head during live solving. I would greatly appreciate your take on this problem.

Side question: how does one get $abc = 1$?

  • $\begingroup$ Can you update your post with what you've tried and where you're stuck? You also mention $x,y,z$, but there are only $a,b,c$ $\endgroup$ – David P Mar 7 '20 at 9:24
  • $\begingroup$ Worthy note: the only solution really is only $a=b=c$, is there any neat way to prove it? $\endgroup$ – Danjel Mar 7 '20 at 9:52

If $a=0$, we get immediately that $a=b=c=0$. Now assume that $abc\neq 0$. The first equation can be written as:


Write the other equations similarly and multiply them:


or $abc(abc-1)(a-b)(b-c)(c-a)=0$. $abc\neq 0$ and if $a=b$ this leads immediately to $a=b=c$. The only case that may lead to different solutions is $abc=1$. In this case the system is equivalent with:

$$\begin{cases}b-c=c^2(c-a)\\ c-a=a^2(a-b)\\ a-b=b^2(b-c)\end{cases}$$

Notice that this implies the $a-b, b-c$ and $c-a$ have the same sign. However since their sum is $0$, this is only possible if $a-b=b-c=c-a=0$ and thus $a=b=c=1$.

In conclusion the only solution is $a=b=c$.


Summing both sides



This leads to one of the solutions


But there is a case where

$a^2(c-1)\neq ab(c-1)$

$b^2(a-1)\neq bc(a-1)$

$c^2(b-1)\neq ac(b-1)$

Now WLOG, Let $a>b>c>0$

Then taking the first equation


Now, Comparing LHS terms with RHS terms

$ab^2>abc ; b>c$

$ac>c^2 ; a>c$

Since the LHS is strictly greater than the RHS, we conclude that

$a\neq b\neq c$

produces no solutions.

  • $\begingroup$ I'm sorry, but how does one get $a=b=c$ from the equation? $\endgroup$ – Danjel Mar 7 '20 at 9:42
  • $\begingroup$ Compare coefficients of $(a-1),(b-1),(c-1)$ $\endgroup$ – h-squared Mar 7 '20 at 9:42
  • $\begingroup$ @h-squared No, it doesn't work that way. $\endgroup$ – Calvin Lin Mar 7 '20 at 9:49
  • $\begingroup$ It is one of the solutions $\endgroup$ – h-squared Mar 7 '20 at 9:50
  • 2
    $\begingroup$ @h-squared, you cannot let WLOG $a>b>c$ because the system is cyclic, not symmetric. You also have to discuss $a<b<c$. $\endgroup$ – LHF Mar 7 '20 at 10:09

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