Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find all positive real numbers $x,y,z$ which satisfy the following equations simultaneously. $x^3+y^3+z^3=x+y+z$

share|cite|improve this question
Hi, and welcome! Can you please share your thoughts on the problem, and explain what's giving you difficulty? This will help people write responses that are appropriate to your question. – user61527 Nov 9 '13 at 2:34
I am unable to find the values of x,y,z – lokesh Nov 9 '13 at 2:38
Can you share what you've tried doing? Can you give some context for the problem, and perhaps mention some techniques you've seen or studied for solving systems like this? – user61527 Nov 9 '13 at 2:38
Don't waste your time with those that ask you those questions. Wait, and someone will answer you for sure. – Mlazhinka Shung Gronzalez LeWy Nov 9 '13 at 2:58
@ABC LOL. Sadly, that's so true. – Calvin Lin Nov 9 '13 at 3:10

The equations do not have positive solutions. It follows directly from AM-GM inequalities.

Recall that $x,y,z> 0$ then


$3(x^2+y^2+z^2)-(x+y+z)^2=(x-y)^2+(y-z)^2+(z-x)^2\geq 0$

Hence $x+y+z\geq (x+y+z)^2$

then $x+y+z\leq 1$.

But $x,y,z>0$, so $x,y,z$ are strictly smaller than 1.

So $x^3<x$, $y^3<y$, $z^3<z$, which means $x^3+y^3+z^3<x+y+z$, a contradiction!

share|cite|improve this answer

This is an answer that I am writing to give an alternative, brainless, path for these kind of problems, as an answer using means will soon appear (has already appeared).

Check the details as I have to go now.

First we use Newton's identities to write everything in terms of the elementary symmetric polynomials.


$e_1=x+y+z$, $p_1=x+y+z$

$e_2=xy+xz+yz$, $p_2=x^2+y^2+z^2$

$e_3=xyz$, $x^3+y^3+z^3$.

Then $e_2=(e_1^2-p_2)/2$

The given equations are $p_3=e_1$ and $p_2=e_3$. So, $e_2=(e_1^2-e_3)/2$

So $Q(t):=(t-x)(t-y)(t-z)=t^3-e_1t^2+\frac{(e_1^2-e_3)}{2}t-e_3$

We also have $p_3=e_1p_2-e_2p_1-3e_3=e_1e_3-\frac{(e_1^2-e_3)}{2}e_1-3e_3$. So, from the given equation, we get $e_1e_3-\frac{(e_1^2-e_3)}{2}e_1-3e_3=e_1$, from where we can solve for $e_3$ to get $e_3=\frac{e_1^3/2}{e_1-e_1/2-3}$.

Then $$Q(t)=t^3-e_1t^2+\frac{(e_1^2-\frac{e_1^3/2}{e_1-e_1/2-3})}{2}t-\frac{e_1^3/2}{e_1-e_1/2-3}$$

Or $$Q(t)=t^3-e_1t^2+\frac{(\frac{-3e_1^2}{e_1/2-3})}{2}t-\frac{e_1^3/2}{e_1/2-3}$$

Now, we try to impose that this polynomial has three positive real roots.

For example: From Descartes' rule of signs, if $e_1/2-3\geq0$ then $Q$ has only one positive root, but it must have 3 or them. So $e_1<3/2$.

We can now compute the discriminant and impose the condition that the roots are reals.

share|cite|improve this answer

x=y=z=0 is the only real solution but not sure if that counts as it is not positive.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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