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!

  • $\begingroup$ Were you given this as an exercise? $\endgroup$ – user117644 Mar 8 '15 at 10:15
  • $\begingroup$ Self-study - from the book "Elementary mathematics" by Dorofeev, Potapov and Rozov. $\endgroup$ – geomquestion Mar 8 '15 at 10:16
  • 1
    $\begingroup$ @mathlove, you should convert that into an answer. It's much simpler than the cubic equations in the other answers. $\endgroup$ – Foo Bar Mar 8 '15 at 15:34
  • $\begingroup$ @FooBar: OK, I did so. $\endgroup$ – mathlove Mar 8 '15 at 21:43
  • $\begingroup$ Related but different: math.stackexchange.com/questions/396915 $\endgroup$ – Watson Dec 24 '16 at 16:24

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


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.


  • $\begingroup$ Nice!!!!!!!!!!! $\endgroup$ – barak manos Mar 8 '15 at 10:33
  • $\begingroup$ In general, when formatting multi-line sequences of equalities, you want the equals signs to be at the start of each line, not the end. $\endgroup$ – David Richerby Mar 8 '15 at 15:18

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 $2$.


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 $$x=2$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.