Cubic factorizations

let $a=\sqrt[3]{9+\sqrt{80}}$ , $b=\sqrt[3]{9-\sqrt{80}}$

then $ab=\sqrt[3]{(9+\sqrt{80})(9-\sqrt{80})} = \sqrt[3]{81-80} = 1$ , and $x = a + b +1$

$x^2(x-3) = x^3-3x^2+3x-1-3x+1 = (x-1)^3-3x+1 = (a+b)^3-3(a+b) - 2 = a^3+3ab(a+b)+b^3-3(a+b)-2 = a^3+b^3-2= 9+\sqrt{80}+9-\sqrt{80}-2 = 16$

Let $\sqrt[3]{9+\sqrt{80}} = a$ and $\sqrt[3]{9+\sqrt{80}} = b$ . Then

$\begin{aligned} (a+b)(a^2 - \blue{ab} + b^2) & = a^3 + b^3 & \small \blue{\text{Note that }ab = \sqrt{(9+\sqrt{80})(9-\sqrt{80})}=\sqrt{81-80}=1} \\ (a+b)(\red{a^2 + b^2} - \blue 1) & = 9+\sqrt{80} + 9 - \sqrt{80} \\ (a+b)(\red{(a+b)^2 - 2ab} - 1) & = 18 \\ (a+b)((a+b)^2 - 3) & = 18 \\ (a+b)^3 - 3(a+b) & = 18 \\ \implies a + b & = 3 \end{aligned}$

Therefore, $X=a+b+1 = 4 \implies X^2(X-3) = 4^2(4-3) = \boxed{16}$ .