Symmetric polynomials with Vieta's and Newton's theorems 
Let $ x_{1}, x_{2}, x_{3}$  be the solutions of the equation $ x^3 -3x^2 + x - 1 = 0.$ 
Determine the values of $$\frac{1}{{x_{1}x_{2}}} + \frac{1}{{x_{2}x_{3}}} + \frac{1}{{x_{3}x_{1}}}$$ 
and also 
$$ x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$$ 


We have
$$x^3-3x^2+x-1=(x-x_1)(x-x_2)(x-x_3)$$
so by developing  the RHS we see that $\sum x_i x_j=1:$ the coefficient of $x$ and similarly  $x_1x_2x_3=1$ and $x_1+x_2+x_3=3$.
Denote $V_1$ the first desired value than $V_1\times(x_1x_2x_3)=x_1+x_2+x_3$. And denote $V_2$ the second value then
$$(x_1+x_2+x_3)^3=V_2+3(x_1x_2+x_2x_3+x_3x_1)(x_1+x_2+x_3)-3x_1x_2x_3$$
