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


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$\#2$:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$a^2+b^2+c^2-ab-bc-ca=(a+b+c)^2-3(ab+bc+ca)$$
