Find rational numbers $\alpha $ and $\beta$ in $\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$ How should we find two rational numbers $\alpha$ and $\beta$ such that
$$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$$
The answer I got $\alpha = 1 $ and $\beta = 1$. If I'm wrong, please correct me. Thank you

You are right.
We can get your answer by the following way.
Let $\sqrt[3]{7+5\sqrt2}=x$.
Thus, $x>2$ and $$x^3=7+5\sqrt2,$$ which gives
$$(x^3-7)^2=50$$ or
$$x^6-14x^3-1=0$$ or
$$x^3-\frac{1}{x^3}-14=0.$$
Now,let $x-\frac{1}{x}=t$.
Thus, $t>0$ and we obtain:
$$t^3+3t-14=0$$ or
$$t^3-2t^2+2t^2-4t+7t-14=0$$ or
$$(t-2)(t^2+2t+7)=0,$$ which gives $$t=2,$$
$$x-\frac{1}{x}=2$$ or
$$x^2-2x-1=0$$ or
$$x=1+\sqrt2,$$ which says $$\alpha=\beta=1.$$
