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Tell me please, how calculate this expression:

$$ \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} $$

The result should be a number.

I try this:

$$ \frac{\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)}{\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)} = $$

$$ = \frac{2 + \sqrt{5} + 2 - \sqrt{5}}{\sqrt[3]{\left(2 + \sqrt{5}\right)^2} + 1 + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}} $$

what next?

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marked as duplicate by Watson, Chinnapparaj R, KReiser, Lord Shark the Unknown, max_zorn Nov 23 '18 at 5:14

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  • $\begingroup$ That made me laugh: "the result should be a number". What else? :D $\endgroup$ – Jack D'Aurizio Sep 1 '15 at 12:30
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Let $s=a+b$ be our sum, where $a=\sqrt[3]{2+\sqrt{5}}$ and $b=\sqrt[3]{2-\sqrt{5}}$. Note that $$s^3=a^3+b^3+3ab(a+b)=a^3+b^3+3abs.$$ Thus since $a^3+b^3=4$ and $ab=\sqrt[3]{-1}=-1$, we have $s^3=4-3s$. This has the obvious root $s=1$ and no other real root.

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    $\begingroup$ It might help to see the roots to note that $s^3+3s-4=(s^2+4)(s-1)$. $\endgroup$ – robjohn Sep 1 '15 at 9:28
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    $\begingroup$ @robjohn Actually, it is $s^3+3s-4 = (s-1)(s^2+s+4)$. $\endgroup$ – Ennar Sep 1 '15 at 10:29
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Would it help you to know that

${2\pm\sqrt5}=\left(\frac{1\pm\sqrt5}2\right)^3$ ?

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  • $\begingroup$ More simple, but not obvious solution. Thank you for callback. $\endgroup$ – Yura Sep 1 '15 at 7:20
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$(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} )^3 \\ =(\sqrt[3]{2 + \sqrt{5}})^3+(\sqrt[3]{2 - \sqrt{5}} )^3+3(\sqrt[3]{2 + \sqrt{5}} ) (\sqrt[3]{2 - \sqrt{5}} )(\sqrt[3]{2 + \sqrt{5}} +\sqrt[3]{2 - \sqrt{5}} ) $

S0 $s^3=4-3s$ From this we get S = 1

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, Let $$x = \sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\;,$$ Then we can write as $$\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}+(-x) = 0$$

Now Using If $$\bullet \; a+b+c = 0\;,$$ Then $$a^3+b^3+c^3 = 3abc$$

So $$\left(2+\sqrt{5}\right)+\left(2-\sqrt{5}\right)-x^3 = 3\left[\sqrt[3]{\left(2+\sqrt{5}\right)\cdot \left(2-\sqrt{5}\right)}\right]\cdot (-x)$$

So $$4-x^3 = -3x\Rightarrow x^3+3x-4=0\Rightarrow (x-1)\cdot (x^2+x+4)=0$$

So we get $$x=1\Rightarrow \sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}} = 1$$

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    $\begingroup$ Another interesting solution. I noticed one little mistake. Instead $(x-1)\cdot (x^2 + 4)=0$ must be $(x-1)\cdot (x^2+ x + 4)=0$. Thanks. $\endgroup$ – Yura Sep 1 '15 at 10:26
  • $\begingroup$ Sorry Yura, I have edited it. $\endgroup$ – juantheron Sep 1 '15 at 10:48

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