# Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ [duplicate]

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?

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

• That made me laugh: "the result should be a number". What else? :D – Jack D'Aurizio Sep 1 '15 at 12:30

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.

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

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

• More simple, but not obvious solution. Thank you for callback. – Yura Sep 1 '15 at 7:20

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

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

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

• 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. – Yura Sep 1 '15 at 10:26
• Sorry Yura, I have edited it. – juantheron Sep 1 '15 at 10:48