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$\begin{aligned} N &= \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \\ N^3 &= \left (\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right )^3 \\ N^3 &= (2 + \sqrt{5}) + (2 - \sqrt{5}) + 3\sqrt[3]{(2 + \sqrt{5})(2 - \sqrt{5})} \left (\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} \right ) \\ N^3 &= 4 + 3 \sqrt[3]{-1} N \\ N^3 &= 4 - 3N \\ N^3 + 3N - 4 &= 0 \\ (N - 1)(N^2 + N + 4) &= 0. \end{aligned}$

Since the discriminant of $x^2 + x + 4$ is $1^2 - 4(4)(1) = -15 < 0,$ there are no real solutions to $N^2 + N + 4 = 0.$ Thus, we conclude that $\boxed{N = 1}.$

[This is not a complete solution. It is similar to Steven's.]

Let $x = \sqrt[3]{2 + \sqrt{5} } , y = \sqrt[3] { 2 - \sqrt{5 } }$ . We get that $x^3 + y^3 = 4$ and $xy = -1$ . Hence, show that $x + y = 1$ .

Solution 1

The problem immediately reminded me of a solution to a cubic equation: $x^3 + px + q = 0 \ \ \ \implies\ \ \ \ x = \sqrt[3]{-\frac12q + \frac12\sqrt{q^2 + \frac4{27}p^3}} + \sqrt[3]{-\frac12q - \frac12\sqrt{q^2 + \frac4{27}p^3}}.$ (This is the real solution if there exists only one.)

Matching this with the expression given here, $-\frac12q = 2 \ \ \ \ \therefore\ \ \ \ q = -4;$ $(-4)^2 + \frac4{27}p^3 = 2^2\cdot 5\ \ \ \ \therefore\ \ \ \ p = 3.$ Thus the expression is the only real solution of the equation $x^3 + 3x - 4 = 0.$ Clearly, $x = 1$ is a real solution of this equation, so that the expression is indeed equal to 1.

Solution 2

Let $a_\pm = 2 \pm \sqrt 5$ . I don't like the look of roots inside roots, so I played around a bit to get rid of that. For instance, $a_{+} a_{-} = (2 + \sqrt 5)(2 - \sqrt 5) = 4 - 5 = -1,$ $\sqrt[3]{a_{+}} \sqrt[3]{a_{-}} = \sqrt[3]{a_{+} a_{-}} = \sqrt[3]{-1} = -1.$

Suppose now that we have two numbers, $x_+$ and $x_-$ , for which this property is true and your calculator is right. $\begin{cases} x_{+} + x_{-} = 1 \\ x_{+} x_{-} = -1.\end{cases}$ They must be the solutions of $x^2 - x - 1 = 0$ , namely $x_\pm = \tfrac12 \pm \tfrac12\sqrt 5.$ Now use $x^2 = x + 1$ to evaluate the cube of either solution: $x^3 = x(x^2) = x(x + 1) = x^2 + x = (x + 1) + x = 2x + 1,$ so that $x_\pm^3 = 2(\tfrac12 \pm \tfrac12\sqrt 5) + 1 = 2 \pm \sqrt 5.$ We see that our $x_\pm$ are precisely the $\sqrt[3]{a_\pm}$ , as we needed to show.

One liner? $\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}= \sqrt[3]{(1/2+1/2\sqrt{5})^3} +\sqrt[3]{(1/2-1/2\sqrt{5})^3} =1$

$2+\sqrt{5}=\frac{1+3\sqrt{5}+3(√5)^{2}+(\sqrt{5})^{3}}{8}$ = $\frac{1+\sqrt{5}}{2}$ ^3

$2-\sqrt{5}=\frac{1+3(-\sqrt{5})+3(-\sqrt{5})^{2}+(-√5)^{3}}{8}$ = $\frac{1-\sqrt{5}}{2}$ ^3

Hence

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

I only noticed that the two terms are golden ratios and they always add to 1. i.e. 1.618..... And -0.618...