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

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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
$$p_3=64-3\cdot4\cdot(-4)+3\cdot(-8)=88\ .$$
