WMC 2018 Senior Problem 26

We can use the cubic formula and "work backwards." We know that the cubic equation $x^3-3px+2q=0$ with positive discriminant $D=q^2-p^3$ has the unique real solution $x=\sqrt[3]{-q+\sqrt{D}}+\sqrt[3]{-q+\sqrt{D}}$ . Here, $q=-7,D=22$ and $p=\sqrt[3]{q^2-D}=3$ , and the answer is $x^3-3px+2q=\boxed{x^3-9x-14}$ .

Let $a=\sqrt[3]{7+\sqrt{22}}$ and $b=\sqrt[3]{7-\sqrt{22}}$ . Consider the following identity:

$\begin{aligned} (a+b)\left(a^2-ab+b^2\right) & = a^3+b^3 \\ {\color{#3D99F6}(a+b)}\left({\color{#3D99F6}(a+b)}^2-3\color{#D61F06}ab\right) & = 7+\sqrt{22}+7-\sqrt{22} & \small \color{#3D99F6} \text{Note that }x=a+b \\ {\color{#3D99F6}x}\left({\color{#3D99F6}x}^2-3\color{#D61F06}(3)\right) & = 14 & \small \color{#D61F06} \text{and }ab = \sqrt[3]{(7+\sqrt{22})(7-\sqrt{22})} = 3 \\ \implies x^3 - 9x & = 14 \end{aligned}$

Therefore, the answer is $\boxed {x^3-9x-14}$ .