A simple answer to a complicated $x$

Let $\alpha = \sqrt[3]{7+\sqrt{22}}$ and $\beta = \sqrt[3]{7-\sqrt{22}}$ , then $x = \alpha+\beta$ .

$\begin{aligned} x^3 + bx + c & = 0 \\ \alpha^3 + 3\alpha^2 \beta + 3\alpha \beta^2 + \beta^3 + bx + c & = 0 \\ 7+\sqrt{22} + 3\alpha \beta ( \alpha + \beta) + 7- \sqrt{22} + bx +c & = 0 \\ \color{#3D99F6}{14} + \color{#D61F06}{3\alpha \beta} x + \color{#D61F06}{b}x + \color{#3D99F6}{c} & = 0 \end{aligned}$

$\Rightarrow \begin{cases} b & = -3\alpha \beta = -3 \left(\sqrt[3]{7+\sqrt{22}}\right) \left( \sqrt[3]{7-\sqrt{22}}\right) \\ & = -3 \sqrt[3]{49-22} = -3 \sqrt[3]{27} = -9 \\ \\ c & = -14 \end{cases}$

$\Rightarrow -bc = - (-9)(-14) = \boxed{-126}$ .

Cube both sides:

$x^3=7+\sqrt{22}+3\sqrt[3]{(7+\sqrt{22})(7-\sqrt{22})}(\sqrt[3]{7+\sqrt{22}}+\sqrt[3]{7-\sqrt{22}})+7-\sqrt{22} \\ x^3=14+3\sqrt[3]{49-22}x \\ x^3=14+9x \\ x^3-9x-14=0 \\ \implies b=-9 \implies c=-14 \implies -bc=\boxed{-126}$

We can set $x$ as $y + z$ . We can easily see that $yz$ = $3$ and $y^3 + z^3 = 14$ . Expanding $x^3$ gives $(y+z)^3$ which is $(y^3 + z^3 + 3(y^2*z + yz^2)) = ((y^3 + z^3 + 3yz(z + y))$ . Now we can substitute values, which gives us $14 + 9(y+z)$ . The whole expression is now: $14 + 9(y+z) + b(y+z) + c$ . Clearly, we can see to cancel out the terms, $b = 9$ and $c = 14$ . So, $b*c = -126$ .