How to show this equals 1 without "calculations"

Solution 1:

$u^3=2+\sqrt 5$, $v^3=2-\sqrt 5$

$u^3v^3=4-5=-1$ so $uv=-1$.

$u^3+v^3=4$

$(u+v)^3=u^3+v^3+3uv(u+v)=4-3(u+v)$.

That is, $u+v$ is a root of the equation $X^3+3X-4=0$ (1). The derivative of this polynomial is $3X^2+3$ which has no roots. By Rolle's theorem, that means that (1) has only one real root. You can easily check that $1$ is that root.

Solution 2:

I'll expand on Guillermo's comment: let $$ a=\sqrt[3]{2+\sqrt{5}},\quad b=\sqrt[3]{2-\sqrt{5}}. $$ You want to show that $a+b=1$. Consider $$ (a+b)^3=a^3+3a^2b+3ab^3+b^3=(a^3+b^3)+3ab(a+b)=4+3\times(-1)(a+b) $$ so if you let $x=(a+b)$ , then $x$ is a real number satisfying $$ 0=x^3+3x-4=x^3-x^2+x^2-x+4x-4\\ =(x-1)(x^2+x+4)=(x-1)\left[(x+0.5)^2+3.75\right]. $$ We infer that $x=1$.

Solution 3:

Let $a=\sqrt[3]{2+\sqrt{5}}$ $b=\sqrt[3]{2-\sqrt{5}}$. Obviously $a b =-1$. Now let's forget what values of $a$ and $b$ and solve the system: $$ ab=-1\\ a+b=1 $$ They imply $$ a^2-a-1=0 $$ namely $a=\frac {1\pm \sqrt{5}}{2}$. So $a=\frac {1+\sqrt{5}}{2}, b=\frac {1-\sqrt{5}}{2}$. Now taking $a^3,b^3$ we can easily get $2\pm\sqrt{5}$.