Sum of two irrational numbers

Question

Prove that this sum of two irrational numbers is rational number $$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}$$

Answer 1

Hint: $2$ is the third Fibonacci number. Using the closed form expression for Fibonacci numbers will help here.

Expanding on hint as requested: The closed form for the fibonacci numbers is $$F_n=\frac{\phi^n-\psi^n}{\sqrt{5}}$$ for $\phi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$. In particular, to get $F_3=2$, you compute $$ \frac{\phi^3-\psi^3}{\sqrt{5}}=\frac{2+\sqrt{5}-(2-\sqrt{5})}{\sqrt{5}}=2. $$ I.e. the cube roots from the question can actually be computed; they are $\phi$ and $\psi$! Taking their sum obviously gives a rational number now as desired.

Answer 2

Call the number $x$. Then show that $x^3=4-3x$. Then solve this equation.

Answer 3

You may notice that $(2-\sqrt{5})(2+\sqrt{5})=-1$, hence the given number is $\alpha-\frac{1}{\alpha}$ with $\alpha$ being a root of $p(x)=(x^3-2)^2-5=x^6-4x^3-1$. On the other hand $p(x)=0$ implies $$x^3-\frac{1}{x^3} = 4$$ which can be written as $$\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)-4=0$$ so the given number is a root of $q(x)=x^3+3x-4 = (x-1)(x^2+x+4)$. Since $x^2+x+4$ has a negative discriminant and $\alpha-\frac{1}{\alpha}$ is real, $\alpha-\frac{1}{\alpha}=\color{red}{1}$.