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 
$$ x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$$ 

 A: Given $(x-x_1)(x-x_2)(x-x_3)=x^3-3x^2+x-1$, the elementary symmetric polynomials of $x_1,x_2,x_3$ are given by Vieta's formulas:
$$ e_1=x_1+x_2+x_3=3,\quad e_2=x_1 x_2+x_1 x_3+x_2 x_3 = 1,\quad e_3=x_1 x_2 x_3 = 1 \tag{1}$$
hence
$$ \frac{1}{x_1 x_2}+\frac{1}{x_1 x_3}+\frac{1}{x_2 x_3} = \frac{e_1}{e_3} = \color{red}{3}.\tag{2}$$
Given the elementary symmetric polynomials, the power sums are given by Newton's identities:
$$ p_1=x_1+x_2+x_3 = 3,\quad p_2=x_1^2+x_2^2+x_3^2=e_1^2-2e_2 = 7,\tag{3} $$
and since $x_j^3 = 3x_j^2-x_j+1$ for every $j\in\{1,2,3\}$,
$$ p_3 = 3p_2-p_1+3 = 21-3+3 = \color{red}{21}.\tag{4}$$
A: For the first case, using Vieta's Formula $$x_1x_2x_3=1\implies\dfrac1{x_1x_2}=x_3$$
For the second, 
Method$\#1$: observe that $$x_r^3-3x_r^2+x_r-1=0\implies x_r^3=3x_r^2-x_r+1\ \ \ \ (r)$$ for $r=1,2,3$
$(1)+(2)+(3)=?$
Now $(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2(x_1x_2+x_2x_3+x_3x_1)$
Method$\#2$:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$a^2+b^2+c^2-ab-bc-ca=(a+b+c)^2-3(ab+bc+ca)$$
A: 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
$$(x_1+x_2+x_3)^3=V_2+3(x_1x_2+x_2x_3+x_3x_1)(x_1+x_2+x_3)-3x_1x_2x_3$$
