How to arrive at Ramanujan's nested radicals?

Question

Ramanujan found that $\sqrt[3]{\cos\left(\frac {2\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {4\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {8\pi}{7}\right)}=\sqrt[3]{\frac {1}{2}\left(5-3\sqrt[3]{7}\right)}$ on the last page of Ramanujan's second notebook.

He also found that $\sqrt[3]{\sec\left(\frac {2\pi}{9}\right)}+\sqrt[3]{\sec\left(\frac {4\pi}{9}\right)}-\sqrt[3]{\sec\left(\frac {\pi}{9}\right)}=\sqrt[3]{6\sqrt[3]{9}-6}$

My question: How do we find the denesting with only the knowledge on the nested radical? Example: How would you find the denestings for $\sqrt[3]{\frac {1}{2}\left(5-3\sqrt{7}\right)}$ without knowledge on $\sqrt[3]{\cos\left(\frac {2\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {4\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {8\pi}{7}\right)}$?

Answer 1

Those identities are not difficult to prove once established. For instance, $\alpha=\cos\left(\frac{2\pi}{7}\right),\beta=\cos\left(\frac{4\pi}{7}\right),\gamma=\cos\left(\frac{6\pi}{7}\right)=\cos\left(\frac{8\pi}{7}\right)$ are algebraic conjugates, roots of the polynomial: $$ p(x) = 8x^3+4x^2-4x-1.$$ If follows that $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$ are roots of the polynomial: $$ q(x) = 8x^9-4x^6-4x^3-1 $$ and it is not difficult to compute, from $q(x)$, a polynomial that vanishes over $\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}$.
With some god-like inspiration, one may notice that $$ r(x) = 4x^9-30x^6+75x^3+32 $$ is a factor of such polynomial. Then the conjecture $$ \alpha^{1/3}+\beta^{1/3}+\gamma^{1/3} = \sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}} $$ is natural, since $r(x)$ is the minimal polynomial of the RHS, and it is straightforward to check.
However, how Ramanujan actually saw that is a mystery to me. Probably the term genius was forged for a reason.

Answer 2

Let's see if I can do something similar to Jack's answer for the second one. I'm not Ramanujan, but I have Maple.

$\sec(2\pi/9), \sec(4\pi/9), \sec(8\pi/9)$ are roots of $t^3 - 6 t^2 + 8$. If $\alpha$, $\beta$, $\gamma$ are cube roots of these, $\alpha^{1/3} + \beta^{1/3} + \gamma^{1/3}$ will satisfy the following polynomial:

factor(evala(Norm(add(RootOf(_Z^3-RootOf(t^3-6*t^2+8,t,index=i)),i=1..3)-z)));

$$- \left( {z}^{18}-72\,{z}^{15}+2484\,{z}^{12}-19008\,{z}^{9}+31104\,{z }^{6}+46656\,{z}^{3}+46656 \right) ^{6} \left( {z}^{9}+18\,{z}^{6}+108 \,{z}^{3}-1728 \right) ^{6}{z}^{18} \left( {z}^{9}-162\,{z}^{6}+157464 \right) \left( {z}^{18}+972\,{z}^{12}+69984\,{z}^{6}+1259712 \right) ^{3} \left( {z}^{27}-162\,{z}^{24}+9612\,{z}^{21}-480384\,{z} ^{18}+16275168\,{z}^{15}-460269216\,{z}^{12}+7616887488\,{z}^{9}- 85571676288\,{z}^{6}+116784820224\,{z}^{3}+1833767424 \right) ^{3} \left( {z}^{27}-162\,{z}^{24}+9612\,{z}^{21}-141696\,{z}^{18}-2667168 \,{z}^{15}-131935392\,{z}^{12}+687911616\,{z}^{9}+28348838784\,{z}^{6} +193536552960\,{z}^{3}+145199817216 \right) ^{3} \left( {z}^{27}-108\, {z}^{21}-6048\,{z}^{18}-54432\,{z}^{15}-54432\,{z}^{12}-451008\,{z}^{9 }+1399680\,{z}^{6}-559872\,{z}^{3}-373248 \right) ^{6} \left( {z}^{27} -108\,{z}^{21}+6048\,{z}^{18}-54432\,{z}^{15}+54432\,{z}^{12}-451008\, {z}^{9}-1399680\,{z}^{6}-559872\,{z}^{3}+373248 \right) ^{6} $$

The factor that is $0$ for the sum of the real cube roots is $$ z^9+18 z^6+108 z^3-1728$$

and the real root of that is $\sqrt[3]{-6 + 6 \sqrt[3]{9}}$.

I don't see how you could go in the reverse direction, starting with $\sqrt[3]{-6 + 6 \sqrt[3]{9}}$ and finding a representation for it (and I doubt that Ramanujan did it that way either).

EDIT: In the same vein:

$$ \cos\left(\frac{2\pi}{9}\right)^{1/3} + \cos\left(\frac{4\pi}{9}\right)^{1/3} + \cos\left(\frac{8\pi}{9}\right)^{1/3} = \frac{\sqrt [3]{-24+12\,\sqrt[3]{9}}}{2}$$