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Solve the following system of equations in $x,y,z \in \Bbb R$: $$ (x+y)^3 = z $$ $$ (x+z)^3 = y$$ $$ (y+z)^3 = x$$

I found the solutions $(0,0,0)$, $(\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})$ and $(-\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}})$, but I'm not sure if my proof is correct. Also I was wondering if there is a more elegant solution than the one I found (described below).

First, I took the first two equations and solved for $x$, getting $$ x = \sqrt[3]{z} - y$$ $$ x = \sqrt[3]{y} - z$$

Then substituting $z = c^3$ and $y = b^3$ (since we are working in $\Bbb R$, these values of $b$ and $c$ exist and are unique), we get

$$c - b^3 = b - c^3$$

and thus

$$c + c^3 = b + b^3$$

Since the function $f(x) = x^3 + x$ is one-to-one and onto, this implies $b=c$. By symmetry, this means $a=b=c$. Since $f(x) = x^3$ is one-to-one and onto, this implies $x=y=z$. Plugging this into the original equation gives

$$8x^3 = x$$

which has the solutions $x = 0, \pm \frac{1}{2\sqrt{2}}$. Hence the three solution pairs I found.

Is this solution ok? Is there an alternate, more elegant approach (that perhap more obviously uses the symmetry between $x$, $y$ and $z$?)

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  • $\begingroup$ Your third equation is wrong, don't? $\endgroup$ – MonsieurGalois Dec 8 '16 at 7:58
  • $\begingroup$ @MonsieurGalois Sorry, that was a typo, you're right. Fixed. $\endgroup$ – aras Dec 8 '16 at 8:01
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This problem is very pretty:

Let's suppose that $x\geq y\geq z$ (any order will let you the same, you can check that). Because of this,

$$x+y\geq x+z\geq z+y$$ $$\Rightarrow (x+y)^3\geq (x+z)^3\geq (z+y)^3$$

But that means that:

$$z\geq y\geq x$$

So from that $x=y=z$ then for any equation you will have $8x^3=x$. From this last equation you have $x=0$ is solution or $x=\pm \frac{1}{2\sqrt{2}}$

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    $\begingroup$ This is awesome, thank you! I knew there would be a better argument...will accept when the accept cooldown wears off. $\endgroup$ – aras Dec 8 '16 at 8:03

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