Finding the cube root

Suppose that $\sqrt[3]{7+5\sqrt2}-\sqrt[3]{7-5\sqrt2}=x$ Cubing both the sides we will get the following equation $x^3+3x-14=0$
By observing we find $2$ as a solution, the other two solutions being non-real get rejected and $2$ is the final answer.

7+5sqrt(2)=[1+sqrt(2)]^3 and 7-5sqrt(2)=[1-sqrt(2)]^3