Let: .$\displaystyle A \,=\,9 + \sqrt{80},\;B \,=\,9-\sqrt{80}$
Note that: .$\displaystyle A + B \:=\: (9+\sqrt{80}) + (9-\sqrt{80}) \:=\:18$
. . . . . . . . . . . $\displaystyle AB \:=\:(9+\sqrt{80})(9 - \sqrt{80}) \:=\:1$
We have: .$\displaystyle X \:=\:A^{\frac{1}{3}} + B^{\frac{1}{3}}$
Cube both sides:
. . $\displaystyle X^3 \;=\;\left(A^{\frac{1}{3}} + B^{\frac{1}{3}}\right)^3$
. . $\displaystyle X^3 \;=\;A + 3A^{\frac{2}{3}}B^{\frac{1}{3}} + 3A^{\frac{1}{3}}B^{\frac{2}{3}} + B$
. . $\displaystyle X^3 \;=\;(A + B) + 3A^{\frac{1}{3}}B^{\frac{1}{3}}\left(A^{\frac{1}{3 }} + B^{\frac{1}{3}}\right) $
. . $\displaystyle X^3 \;=\;\underbrace{(A+B)}_{18} + 3\underbrace{(AB)^{\frac{1}{3}}}_{1^\frac{1}{3}}} \underbrace{\left(A^{\frac{1}{3}} + B^{\frac{1}{3}}\right)}_3$
. . $\displaystyle X^3 \;=\;18 + 3(1)(3)$
. . $\displaystyle X^3 \;=\;27$
Therefore: .$\displaystyle X \;=\;3$