Calculate $\;$ $x^3 + 3x - 14$ $\;$ for $\;$ $x = \sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{1}{\sqrt[3]{7 + 5 \sqrt{2}}}$
$x^3 + 3x - 14$
$= \left(\sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{1}{\sqrt[3]{7 + 5 \sqrt{2}}}\right)^3 + 3 \left(\sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{1}{\sqrt[3]{7 + 5 \sqrt{2}}}\right) - 14$
$= \left(\sqrt[3]{7 + 5 \sqrt{2}}\right)^3 - 3 \times \left(\sqrt[3]{7 + 5 \sqrt{2}}\right)^2 \times \dfrac{1}{\sqrt[3]{7 + 5 \sqrt{2}}}$
$\hspace{1cm}$ $+ 3 \times \sqrt[3]{7 + 5 \sqrt{2}} \times \dfrac{1}{\left(\sqrt[3]{7 + 5 \sqrt{2}}\right)^2} - \left(\dfrac{1}{\sqrt[3]{7 + 5\sqrt{2}}}\right)^3$
$\hspace{3.5cm} + 3 \times \sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{3}{\sqrt[3]{7 + 5 \sqrt{2}}} - 14$
$= 7 + 5 \sqrt{2} - 3 \sqrt[3]{7 + 5 \sqrt{2}} + \dfrac{3}{\sqrt[3]{7 + 5 \sqrt{2}}} - \dfrac{1}{7 + 5 \sqrt{2}}$
$\hspace{3.5cm} + 3 \sqrt[3]{7 + 5 \sqrt{2}} - \dfrac{3}{\sqrt[3]{7 + 5 \sqrt{2}}} - 14$
$= 7 + 5 \sqrt{2} - \dfrac{1}{7 + 5 \sqrt{2}} - 14$
$= 7 + 5 \sqrt{2} - \dfrac{\left(7 - 5 \sqrt{2}\right)}{\left(7 + 5 \sqrt{2}\right) \left(7 - 5 \sqrt{2}\right)} - 14$
$= 7 + 5 \sqrt{2} - \dfrac{\left(7 - 5 \sqrt{2}\right)}{49 - 50} - 14$
$= 7 + 5 \sqrt{2} + 7 - 5 \sqrt{2} - 14$
$= 0$