Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational

Question

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!

Answer 1

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.

Answer 2

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 $$\sqrt[3]{7+5\sqrt 2}+\sqrt[3]{7-5\sqrt 2}=(1+\sqrt 2)+(1-\sqrt 2)=2.$$

Answer 3

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 $2$.

Answer 4

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 $$x=2$$

Answer 5

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.