# Proving Ramanujan's Nested Cube Root

Ramanujan's Nested Cube: If $\alpha,\beta$ and $\gamma$ are roots of the equation$$x^3-ax^2+bx-1=0\tag{1}$$then, for a suitable determination of roots,$$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=(a+6+3t)^{1/3}\tag{2.1}$$and$$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=(b+6+3t)^{1/3}\tag{2.2}$$where$$t^3-3(a+b+3)t-(ab+6(a+b)+9)=0\tag3$$

This identity is what Ramanujan used to get the identity$$\sqrt[3]{\cos\tfrac {2\pi}7}+\sqrt[3]{\cos\tfrac {4\pi}7}+\sqrt[3]{\cos\tfrac {8\pi}7}=\sqrt[3]{\tfrac 12\left(5-3\sqrt[3]7\right)}$$ By starting off with $x^3+x^2-2x-1=0$ and finding the respective roots. Then, by noting that $a=-1,b=-2$, you get the RHS.

Question:

1. How to you prove the formula
2. Is there a standard procedure to find the trigonometric roots of a polynomial

I first started off with a function $x^3-px^2+qx-1=0$ and assumed that the roots were $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$. That way, by Vieta's formula, we have$$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=p$$$$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=-q$$However, I'm not sure how to represent the RHS as $(2.1)$ or $2.2$

EDIT: I found a proof, but something doesn't match up. I have posted another question here

• I am getting $\cos 2\pi/7,\cos 4\pi/7,\cos 8\pi/7$ to be the roots of equation $$8x^3+4x^2-4x-1=0$$ – Navin May 12 '17 at 19:04
• @Pink Yes, that is true. What Ramanujan did was start with$$x^3+x^2-2x-1=0$$To get the roots as$$x=2\cos\frac {2k\pi}{7}$$And then divided by $2$ on both sides. However, if you started with your cubic, you would get the same equality. – Crescendo May 12 '17 at 19:10

Let $$\sqrt[3]{\cos\tfrac {2\pi}7}+\sqrt[3]{\cos\tfrac {4\pi}7}+\sqrt[3]{\cos\tfrac {8\pi}7}=x$$ and $$\sqrt[3]{\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7}+\sqrt[3]{\cos\tfrac {2\pi}7\cos\tfrac {8\pi}7}+\sqrt[3]{\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7}=y.$$ Hence, since $$\cos\tfrac {2\pi}7+\cos\tfrac {4\pi}7+\cos\tfrac {8\pi}7=-\frac{1}{2},$$ $$\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7+\cos\tfrac {2\pi}7\cos\tfrac {8\pi}7+\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7=-\frac{1}{2}$$ and $$\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7=\frac{1}{8},$$ we obtain: $$x^3=\cos\tfrac {2\pi}7+\cos\tfrac {4\pi}7+\cos\tfrac {8\pi}7+3xy-3\sqrt[3]{\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7}=-\frac{1}{2}+3xy-\frac{3}{2}=3xy-2$$ and $$y^3=\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7+\cos\tfrac {2\pi}7\cos\tfrac {8\pi}7+\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7+$$ $$+3xy\sqrt[3]{\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7}-3\sqrt[3]{\cos^2\tfrac {2\pi}7\cos^2\tfrac {4\pi}7\cos^2\tfrac {8\pi}7}=$$ $$=-\frac{1}{2}+\frac{3}{2}xy-\frac{3}{4}=\frac{3}{2}xy-\frac{5}{4}.$$ Thus, $$x^3y^3=(3xy-2)\left(\frac{3}{2}xy-\frac{5}{4}\right)$$ or $$4x^3y^3=18x^2y^2-27xy+10$$ or $$8x^3y^3-36x^2y^2+54xy-20=0$$ or $$(2xy-3)^3+7=0$$ or $$xy=\frac{1}{2}\left(3-\sqrt[3]7\right),$$ which gives $$x^3=\frac{3}{2}\left(3-\sqrt[3]7\right)-2$$ or $$x=\sqrt[3]{\frac{1}{2}\left(5-3\sqrt[3]7\right)}$$ and we are done!
In the general we just need to solve a cubic equation of $xy$.
We have $\alpha$ , $\beta$ and $\gamma$ are roots of the equation $x^3-ax^2+bx-1=0$ so \begin{eqnarray*} a= \sum \alpha \\ b= \sum \alpha \beta \\ \alpha \beta \gamma =1 \end{eqnarray*} so $\sqrt[3]{\alpha \beta \gamma} =1$ and let \begin{eqnarray*} A= \sum \sqrt[3]{\alpha} \\ B= \sum \sqrt[3]{\alpha \beta} \end{eqnarray*} Cube these equations and multiply them \begin{eqnarray*} A^3= a+3\sum \sqrt[3]{\alpha^2 \beta} +6\\ B^3= b+3\sum \sqrt[3]{\alpha^2 \beta} +6\\ AB= \sum \sqrt[3]{\alpha^2 \beta}+3 \end{eqnarray*} let $t=\sum \sqrt[3]{\alpha^2 \beta}$ and cube $AB=t+3$ We have \begin{eqnarray*} t^3+9t^2+27t+27 = A^3B^3=(a+3t +6)(b+3t +6)\\ t^3=3(a+b+3)t+(ab+6(a+b)+9). \end{eqnarray*} Thus the equation is shown.