4

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?

Martin Sleziak
  • 53,687
Yura
  • 117

4 Answers4

11

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.

André Nicolas
  • 507,029
3

Would it help you to know that

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

Senex Ægypti Parvi
  • 2,419
1

$(\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

Way to infinity
  • 1,884
0

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

juantheron
  • 53,015