Find the value below 
Let $a,b,c$ be the roots of the equation $$8x^{3}-4x^{2}-4x+1=0$$
Find $$\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$$

It's just for sharing a new ideas, thanks:)
 A: If $a,b,c$ are the roots of $p(x)=8x^3-4x^2-4x+1$ then $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are the roots of
$$ q(x) = x^3-4x^2-4x+8 $$
so:
$$ \frac{1}{a^3} = \frac{4}{a^2}+\frac{4}{a}- 8 $$
and:
$$\sum_{cyc}\frac{1}{a^3}=4\sum_{cyc}\frac{1}{a^2}+4\sum_{cyc}\frac{1}{a}-24$$
then Viète's theorem applied to $q(x)$ gives:
$$\sum_{cyc}\frac{1}{a^3}=4(4^2-2\cdot(-4))+4(4)-24 = \color{red}{88}.$$

With a bit of experience, one may recognize $p(x)$ as the minimal polynomial of $\alpha=-\cos\frac{2\pi}{7}$, whose conjugates are $-\cos\frac{4\pi}{7}$ and $-\cos\frac{6\pi}{7}$. By this way, the problem is equivalent to proving a not-so-difficult trigonometric identity.

Another neat trick is the following: given $p(x)=\left(1-\frac{x}{a}\right)\left(1-\frac{x}{b}\right)\left(1-\frac{x}{c}\right)$,
$$ \log p(x) = \sum_{cyc}\log\left(1-\frac{x}{a}\right) = -\sum_{cyc}\left(\frac{x}{a}+\frac{x^2}{2a^2}+\frac{x^3}{3a^3}+\ldots\right) $$
hence $\sum_{cyc}\frac{1}{a^3}$ is minus three times the coefficient of $x^3$ in the Taylor series of $\log p(x)$ in a neighbourhood of $x=0$, or:
$$ \sum_{cyc}\frac{1}{a^3} = -\frac{1}{2}\left.\frac{d^3}{dx^3}\log p(x)\right|_{x=0}, $$
and we just need to evaluate:
$$\frac{\left(4+8 x-24 x^2\right)^3}{\left(1-4 x-4 x^2+8 x^3\right)^3}+\frac{3 (-4+24 x) \left(-4-8 x+24 x^2\right)}{\left(1-4 x-4 x^2+8 x^3\right)^2}-\frac{24}{1-4 x-4 x^2+8 x^3} $$
at $x=0$ to get $4^3+3\cdot 4^2-24=\color{red}{88}$ just as before.
A: A systematic way to compute sums of the form $a^{-k} + b^{-k} + c^{-k}$ is
using Newton's identities.
Notice
$$\begin{align}
 a, b, c \text{ roots of } p(x) &= 8x^3 - 4x^2 - 4x + 1\\
\implies\quad
\frac{1}{a}, \frac{1}{b}, \frac{1}{c}
\text{ roots of } q(x) &= x^3 p(\frac{1}{x}) = x^3 - \color{red}{4} x^2 - \color{green}{4}x + \color{blue}{8}
\end{align}
$$
Let $p_k = a^{-k} + b^{-k} + c^{-k}$ for $k = 1,2,3$. Newton's identities tell us:
$$\begin{cases}
p_1 - \color{red}{4} \times 1 &= 0\\
p_2 - \color{red}{4} p_1 - \color{green}{4}\times 2 &= 0\\
p_3 - \color{red}{4} p_2 - \color{green}{4} p_1 + \color{blue}{8}\times 3 &= 0
\end{cases}
\quad\implies\quad
\begin{cases}
p_1 &= 4\\
&\;\Downarrow\\
p_2 &= 4\times \underbrace{4}_{p_1} + 4\times 2 = 24\\
&\;\Downarrow\\
p_3 &= 4\times \underbrace{24}_{p_2} + 4\times \underbrace{4}_{p_1} - 8\times 3 = 88
\end{cases}
$$
A: It has been observed by others that the reciprocals $x_i$ $(1\leq i\leq 3)$ of $a$, $b$, $c$ are the solutions of the equation
$$x^3-4x^2-4x+8=0\ .\tag{1}$$
The power sum $$p_3:=x_1^3+x_2^3+x_3^3={1\over a^3}+{1\over b^3}+{1\over c^3}$$ is a symmetric function of the $x_i$. Therefore it can be expressed as a polynomial in terms of the elementary functions $\sigma_j$ $(1\leq j\leq 3)$ of the $x_i\,$: One easily verifies that
$$x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3\ ,$$
or $$p_3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3\ .$$
Now $(1)$ and Vieta's theorem tell us that
$$\sigma_1=4,\quad \sigma_2=-4,\qquad \sigma_3=-8\ .$$
It follows that
$$p_3=64-3\cdot4\cdot(-4)+3\cdot(-8)=88\ .$$
