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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!

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

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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.

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  • $\begingroup$ 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
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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 $2$.

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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 $$x=2$$

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