Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give me some hints? Thanks in advance!
 A: Let $(a+b\sqrt{2})^3=7+5\sqrt{2}$, then $(a-b\sqrt{2})^3=7-5\sqrt{2}$. This gives us
\begin{align}
a^3+6ab^2 & = 7\\
2b^3+3a^2b & = 5
\end{align} 
From the two equations listed above we get.
$$\frac{a(a^2+6b^2)}{b(2b^2+3a^2)}=\frac{7}{5}.$$
Now let $x=\frac{a}{b}$. Then
$$5x(x^2+6)=7(3x^2+2).$$
Finally we get
$$5x^3-21x^2+30x-14=0$$
Observe that $x=1$ is an obvious solution of this equation. So we can rewrite this as:
$$(x-1)(5x^2-16x+14)=0.$$
The quadratic factor has only non-real roots. Thus $x=1$ is the only real solution. This gives $a=b=1$. Thus the given expression is equal to $2$.
A: First, let us cube the number at hand:
$$\begin{align*}
x^3
&=7+\sqrt{50}+7-5\sqrt{2}+3(7+\sqrt{50})^\frac{2}{3}(7-5\sqrt{2})^\frac{1}{3}+3(7+\sqrt{50})^\frac{1}{3}(7-5\sqrt{2})^\frac{2}{3}\\
&=14+3((49-50)(7+\sqrt{50}))^\frac{1}{3}+3((49-50)(7-5\sqrt{2}))^\frac{1}{3}\\
&=14-3x
\end{align*}$$
Then, $x^3=14-3x \implies x^3+3x-14=0$.
The only real solution of this equation is $x=2$, which is a rational number.
QED.
A: use this well known identity
$$(a+b)^3=a^3+b^3+3ab(a+b)$$,so we Let
$$x=\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}$$
then we have
$$x^3=14+3\sqrt[3]{49-50}\cdot x=-3x+14$$
so $$x=2$$
A: This should be an easy way (I converted my comment into an answer) :
Since $$7±5\sqrt 2=1±3\sqrt2 +6±2\sqrt 2=(1±\sqrt 2)^3,$$
we have
$$\sqrt[3]{7+5\sqrt 2}+\sqrt[3]{7-5\sqrt 2}=(1+\sqrt 2)+(1-\sqrt 2)=2.$$
A: All of the other approaches here are based on guessing that the expressions under the cube root signs are perfect cubes.
Here's an approach that doesn't assume that.
Let $$a = \sqrt[3]{7 + \sqrt{50}} = \sqrt[3]{7 + 5\sqrt{2}}, \quad b = \sqrt[3]{7 - 5\sqrt{2}}.$$
We are trying to evaluate $s = a + b$. We have
$$a^3 = 7 - 5\sqrt{2}, \quad b^3 = 7 + 5\sqrt{2}.$$
So
$$a^3 + b^3 = 14, \quad a^3 b^3 = -1.$$
Since $a$ and $b$ are real, the second equation implies $ab = -1$. Now we find
$$14 = a^3 + b^3 = (a+b)^3 - 3ab(a + b) = s^3 +3s.$$
This cubic equation has only one real solution, $s = 2$, which can be found using the rational root theorem.
