RagvaloD wrote: $a$ is irrational , but $a$ and $a^3-6a$ are roots of square polynomial with integer coefficients.Find $a$ So, for some integers $s,p$ : $a+(a^3-6a)=s$ which is $a^3-5a-s=0$ (e1) $a(a^3-6a)=p$ which is $a^4-6a^2-p=0$ (e2) Multiplying (e1) by $a^2+sa-1$, we get $a^5+sa^4-6a^3-6sa^2+(5-s^2)a+s=0$ Multiplying (e2) by $a+s$, we get $a^5+sa^4-6a^3-6sa^2-pa-ps=0$ Subtracting, we get $(5-s^2+p)a+s+ps=0$ Since $a\notin\mathbb Q$, this implies $s^2=p+5$ and $s(p+1)=0$ If $s=0$, we get $p=-5$ and quadratic $x^2-5=0$ and $x=\pm\sqrt 5$ If $s\ne 0$, we get $p=-1$ and $s^2=4$ and quadratics $x^2\pm 2x-1=0$ Hence the solutions $\boxed{a\in\left\{-1-\sqrt 2,-\sqrt 5,1-\sqrt 2,-1+\sqrt 2,\sqrt 5,1+\sqrt 2\right\}}$