5 . Find the root of the equation $x^{x^{3}}=3$.


5. $\sqrt[3]{3}$. , (Let $x^{3}=y^{\prime}$, we get $x^{y}=3$, $x^{3y}=3^{3}$. That is, $y^{y}=3^{3}$. The equation clearly has a root $y^{\prime}=2$, i.e., $x=\sqrt[3]{3}$. Let's see if there are any other roots: if $y>3$, then $y^{3}>y^{3}>3^{3}$; if $0<y<3$, then $y^{y}<y^{3}<3^{3}$. Therefore, the only root of this problem is $x=\sqrt[3]{3}$,