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How many triplets of real numbers $(x, y, z)$ which satisfy : $$(x + y)^3 = z$$ $$(y + z)^3 = x$$ $$(z + x)^3 = y$$

I need some approaches for solving this problem.

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hint: x=y=z=0 is the only such triplet – Bhargav Aug 27 '11 at 17:48
@Bhargav:How about $x=y=z= \pm \frac{1}{2\sqrt{2}}$? – Quixotic Aug 27 '11 at 17:50
ya, i seem to have missed it – Bhargav Aug 28 '11 at 6:29
up vote 13 down vote accepted

Look first for solutions of the shape $x=y=z=a$. Our equations all reduce to $(2a)^3=a$, which has the solutions $a=0$ and $a=\pm 2^{-3/2}$.

Now look for solutions where not all the variables are equal. For definiteness, look for solutions with $x \lt z$.

If $x \lt z$, then $x+y \lt y+z$ and therefore $(x+y)^3 \lt (y+z)^3$. From the first two equations, it follows that $z \lt x$, which is impossible.

Thus there are $3$ triples that satisfy the system of equations.

One can write up the same idea by starting from $x \le z$ and concluding that $z\le x$, which shows that $x=z$. So for a solution, we must have $x=y=z$.

Other approaches: We sketch a more "algebraic" approach which happens to be more work. From the first two equations, we obtain $$(x+y)^3-(y+z)^3=z-x.$$ Let $Z=x+y$ and $X=y+z$. Factoring the expression $Z^3-X^3$ on the left, we obtain $$(Z-X)(Z^2+ZX+X^2)= (x-z)(Z^2+ZX+X^2)=z-x.$$ If $z \ne x$, this forces $Z^2+ZX+X^2=-1$. But this last equation does not have real solutions. There are many ways to see this, such as the Quadratic Formula. A cuter way is to note that $$4(Z^2 +ZX+X^2)=(2Z+X)^2+3X^2 \ge 0.$$ So we must have $z=x$. Similarly, $z=y$, and we end up looking for solutions of $(2a)^3=a$.

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