Finding two rationals that satisfy an equation So the homework question goes as follows:

Find rational numbers such that $ \sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$

The homework only asks for one pair of $\alpha$ and $\beta$ and it is almost trivial to find the solution $\alpha=\beta = 1$
Solution:
$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$
$7+5\sqrt{2}=(\alpha+\beta\sqrt{2})^3$
$7+5\sqrt{2}=\alpha^3+6\beta^2\alpha +\sqrt{2}(3 \alpha^2 \beta +2\beta^3)$
By identification of coefficients:
$7=\alpha^3+6\beta^2\alpha \space and \space 5=3 \alpha^2 \beta +2\beta^3$
So I just "saw" that $\alpha=\beta = 1$ is a solution
My question is, how do we find other solutions?
Thank you!

The real solutions of a cube root of a positive number are positive numbers, and if you have two positive real-valued solutions then they must be equal. So if $\alpha_1 + \beta_1 \sqrt{2} = \alpha_2 + \beta_2 \sqrt{2}$ are two different solutions where all $\alpha_i, \beta_i$ are rational then you will be able to solve for $\sqrt{2}$ as a rational number, which is impossible. So there is only at most one solution.
