Recall that
[tex](a+b)^3=a^3+3a^2b+3ab^2+b^3[/tex]
which means
[tex]\left(r+\dfrac1r\right)^3=r^3+3r^2\left(\dfrac1r\right)+3r\left(\dfrac1r\right)^2+\left(\dfrac1r\right)^3=r^3+3r+\dfrac3r+\dfrac1{r^3}[/tex]
Given that [tex]r+\frac1r=\sqrt2[/tex], we have
[tex](\sqrt2)^3=r^3+3\sqrt2+\dfrac1{r^3}[/tex]
[tex]\implies r^3+\dfrac1{r^3}=2\sqrt2-3\sqrt2=\boxed{-\sqrt2}[/tex]
