Express this sum of radicals as an integer?

Question

I have read somewhere that the radical $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$ and I don't understand it. How do you solve this(when the RHS is unknown)?

Answer 1

Let us consider the equation $ y^3+py+q=0 $. The theory of its solution tells, that we should consider $$ \Delta=\left(\frac{p}3\right)^3+\left(\frac{q}2\right)^2. $$ If $\Delta>0$, then $$ y=\sqrt[3]{-\frac{q}2-\sqrt{\Delta}}+\sqrt[3]{-\frac{q}2+\sqrt{\Delta}} $$ is unique real solution.

If you know this, the rest is simple. We guess $q=-4$, then $p=3$ and we have equation $$ y^3+3y-4=0 $$ with an obvious real solution $y_0=1$.

Answer 2

Let $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$. Write $a=2+\sqrt{5}$ and $b=2-\sqrt{5}$. Then $$ x^3= a+b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} = a+b + 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) =4-3x $$

Now, the derivative of $x^3+3x-4$ is always positive, which means there is only one real root. By inspection, $x=1$ is that root.

Answer 3

Let $~\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=a.~$ Cubing both sides, using ${\underbrace{(x+y)}_a}^3=x^3+y^3+3xy~\underbrace{(x+y)}_a$, and simplifying, we have $a^3=4-3a$, whose only real solution, $a=1$, can be found quite easily via the rational root theorem. After dividing $a^3+3a-4$ by $a-1$, we're left with the quadratic $a^2+a+4$, whose roots are both complex, but $a\in$ R.