<s>[INST] 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:)
 [/INST] 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:
$$\begin{cases}
p_1 - \color{red}{4} \times 1 &= 0\\
p_2 - \color{red}{4} p_1 - \color{green}{4}\times 2 &= 0\\
p_3 - \color{red}{4} p_2 - \color{green}{4} p_1 + \color{blue}{8}\times 3 &= 0
\end{cases}
\quad\implies\quad
\begin{cases}
p_1 &= 4\\
&\;\Downarrow\\
p_2 &= 4\times \underbrace{4}_{p_1} + 4\times 2 = 24\\
&\;\Downarrow\\
p_3 &= 4\times \underbrace{24}_{p_2} + 4\times \underbrace{4}_{p_1} - 8\times 3 = 88
\end{cases}
$$
</s>