Use Ramanujan’s method to denest $\sqrt[3]{7\sqrt[3]{20}-1}$ and $\sqrt[3]{7\sqrt[3]{20}-19}$ A possible way to denest $(2^{1/3}-1)^{1/3}$ is by first setting $x=\sqrt[3]{2}$ so$$x^3-1=1\implies x-1=\frac 1{1+x+x^2}=\frac 3{1+3x+3x^2+x^3}$$Multiply both sides by $9$ so$$9(x-1)=\left(\frac 3{1+x}\right)^3\implies\sqrt[3]{9(\sqrt[3]2-1)}=1-\sqrt[3]2+\sqrt[3]4$$However, I'm having trouble adapting this method to denest 
both
$$\begin{align*}\sqrt[3]{7\sqrt[3]{20}-1} & =\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}\tag1\\\sqrt[3]{7\sqrt[3]{20}-19} & =\sqrt[3]{\frac 49}-\sqrt[3]{\frac {80}9}+\sqrt[3]{\frac {25}9}\tag2\end{align*}$$
I've only started on (1) so far. Here's my work.

My work: Let $x^3=6860$ so $x^3-1=19^3$. Hence$$x-1=\frac {19^3}{1+x+x^2}=\frac {19^3\cdot3}{3+3x+3x^2}$$However, I'm not quite sure what to do after that. I don't see an easy relationship between three and $x$.

Perhaps you guys can help?

$x^3 +a x^2 +bx +c $ is called a Ramanujan cubic polynomial if
$$b +a c^{1/3} +3c^{2/3}=0\tag1 $$
Its three roots then satisfy
$$\sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 }=\sqrt[3]{ 3\sqrt[3]{9c-ab }-a-6\sqrt[3]c} \tag2$$
Note that (2) contains nested cubic radicals of the form $\sqrt[3]{\sqrt[3] A -B}$ and it can be used to  denest such radicals. Take $\sqrt[3]{7\sqrt[3]{20}-1}$ for example, we set
$$a+6c^{1/3}=1,\>\>\>\>\>27(9c-ab)=7^3\cdot 20$$
Along with (1), we solve for $a=-\frac{37}3$, $b=\frac{340}{27}$,  $c=\frac{8000}{729}$ and the corresponding polynomial
$$x^3 -\frac{37}3x^2+\frac{340}{27}x + \frac{8000}{729}
= (x+\frac59)(x-\frac{16}9)(x-\frac{100}9)$$
from which the Ramanujan’s property (2) yields the denesting
$$
\sqrt[3]{7\sqrt[3]{20}-1} =\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}
$$
Above procedure can be used to denest $\sqrt[3]{7\sqrt[3]{20}-19}$ as well. Note that the method is effective without prior knowledge of results.
