Let $r$ be any real root of $t^5=2t^3+t-2$. Then $(r,r,r)$ is a solution; and these are the only solutions. Note that $t\mapsto t^5$ is strictly increasing (clear). Moreover, we claim $t\mapsto 2t^3+t-2$ is increasing, as well. To see this, set $\varphi(t) = 2t^3+t-2$, note $\varphi'(t) = 6t^2+1>0$. Equipped with this, we are now in position to conclude the proof. Let $x=\max\{x,y,z\}$ without any loss of generality. Then $x\ge z$ implies $x^5\ge z^5$. Consequently, $2y^3+y-2\ge 2x^3+x-2$. Hence $y\ge x$. Thus $x=y$. This, in turn, implies $x=y=z$. Hence, $x,y,z$ are all roots of $t^5=2t^3+t-2$. Remark. The expression above clearly has $1$ as its root. It appears, however, that; two of its other roots are real; whereas the remaining two are complex. What is strange though is that it seems $t^5-2t^3-t+2$ admits no factors in $\mathbb{Q}[t]$ besides $(t-1)$. So I'm curious if it is really asking *all triples* w/o any extra condition.

wording comes from tst-2010-5.pdf from here, no extra conditions mentioned