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:)

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.
