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
 A: 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 $$\alpha=\beta=1.$$
A: From $a^{1/3}=b$ you get $a=b^3$. If $b=(x+y)^3$, then $b^3=x^3+3x^2y+3xy^2+y^3.$
Can you take it from here ?
A: Use The conjugate Radical Roots Theorem
$$7+5\sqrt2=(a+\sqrt2b)^3\iff7-5\sqrt2=?$$
$$a^2-2b^2=-1$$
Compare the rational parts
$$7=a^3+6ab^2=a^3+3a(a^2+1)$$
The only real root is $1$
A: 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, $\alpha=\beta=1$.
A: By Gauss' lemma, if $\alpha$ and $\beta$ are rational, then they are integers. Simply cube both sides to get
$$7+5\sqrt{2}=(\alpha+\beta\sqrt{2})^3.$$
Expanding the right hand side and comparing coefficients shows that
\begin{eqnarray*}
7&=&\alpha^3+6\alpha\beta^2&=&\alpha(\alpha^2+6\beta^2),\\
5&=&3\alpha^2\beta+2\beta^3&=&\beta(3\alpha^2+2\beta^2),
\end{eqnarray*}
so $\alpha$ divides $7$ and $\beta$ divides $5$. This leaves very few options to check.
