<s>[INST] 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
 [/INST] Apply the detesting formula
$$\sqrt[3]{a+b \sqrt R}=\frac12\sqrt[3]{3bs-a}\left(1+\frac1s \sqrt R\right)\tag1$$
where $s$ is the solution of
$$s^3-\frac{3a}b s^2+3R s-\frac{a}b R =0$$
So, for $\sqrt[3]{7+5\sqrt{2}}$, solve for $s$ in
$$s^3-\frac{21}5s^2+6s-\frac{14}5=\frac15(s-1)(5s^2-16s+14)=0$$
which yields $s=1$. Plug into (1) to obtain the denestation
$$\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$$
or, $\alpha=\beta=1$.
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