Denesting Phi, Denesting Cube Roots I have been looking into denesting square roots but I have found that $\sqrt[3]{2+\sqrt{5}}$ equals $(1+\sqrt{5})/2$. The same is true for  $\sqrt[3]{2-\sqrt{5}}$ and $(1-\sqrt{5})/2$. I cannot figure how this is true. I proved this by setting both equal to x and forming polynomial equations. $y= x^2-x-1$ and $y=x^6-4x^3-1$. I do not know how to solve a degree 6 polynomial by hand like that. My objective is to denest the original  $\sqrt[3]{2+\sqrt{5}}$ without knowledge of the simplified phi.
Hint : As a general rule, when dealing with nested radicals of the form $\sqrt[n]{A+B\sqrt[m]C}$ , you write $A+B\sqrt[m]C=(a+b\sqrt[m]C)^n$, and then employ Newton's binomial theorem. In our case, we have

$$\left(a+b\sqrt5\ \right)^3=a^3+3a^2(b\sqrt5)+3a(b\sqrt5)^2+(b\sqrt5)^3=\underbrace{(a^3+15ab^2)}_2+\underbrace{(3a^2b+5b^3)}_1\sqrt5$$ $$\iff a^3+15ab^2=2(3a^2b+5b^3)\iff a^3+15ab^2-6a^2b-10b^3=0\quad|:b^3\iff$$ $$\iff\left(\frac ab\right)^3-6\left(\frac ab\right)^2+15\left(\frac ab\right)-10=0\iff x^3-6x^2+15x-10=0\iff x=1$$ $$\iff\frac ab=1\iff a=b\iff a^3+15a^3=2\iff16a^3=2\iff a=\sqrt[3]\frac18=\frac12$$ or $3b^3+5b^3=1\iff b=\sqrt[3]\frac18=\frac12$ . Similarly for when the second term is $-1$.
It is so easy to solve x6−4x3−1=0
just let x3=t and refine the equation as t2-4t-1=0
after you find the t, you can bring back the x3
for example if t=5, then x3=5 and x=3√5