You need to remember that Vieta's formulas provide the relations between coefficients of equation and its roots.
The problem provides an polynomial equation of third order,`x^3 -3x +1=0` , hence it needs to have three roots.
You should use Vieta's formulas such that:
`x_1 + x_2 + x_3= -0/1` (coefficient of term that contains `x^2` is 0, since this term is missing) => `x_1 + x_2 + x_3 = 0`
`x_1*x_2+x_2*x_3+x_1*x_3 = -3/1 = -3`
`x_1*x_2*x_3 = -1/1 = -1`
You need to evaluate the sum `1/x_1 + 1/x_2 + 1/x_3` , hence you should bring these terms to a common denominator `x_1*x_2*x_3` such that:
`1/x_1 + 1/x_2 + 1/x_3 = (x_2*x_3 + x_1*x_3 + x_1*x_2)/x_1*x_2*x_3`
You need to substitute -1 for `x_1*x_2*x_3` and -3 for `x_1*x_2+x_2*x_3+x_1*x_3` such that:
`1/x_1 + 1/x_2 + 1/x_3 = (-3)/(-1)`
`1/x_1 + 1/x_2 + 1/x_3 = 3`
Hence, evaluating the sum `1/x_1 + 1/x_2 + 1/x_3` yields `1/x_1 + 1/x_2 + 1/x_3 = 3` .
This is a 3rd degree equation.
By theory of equation, the relation between roots and coefficients of the terms of the given equation is as follows:
sum of roots x1+x2+x3 = -coefficient of x^2/coefficient of x^3 =0/1=0
x1x2+x2x3+x3x1 = coefficient of x /coefficient of x^3 = -3/1 =-3 and
x1x2x3 = -constant term/ coefficient of x^3 =-1/1 =-1.
1/x1 +1/x2 +1/x3 = (x2x3+x3x1+x1x2)/(x1x2x3) = -3/(-1) = 3
Transform by 1/x = y or x = 1/y to get the equation whose roots are reciprocal of the given equation.
Then f(1/y) = 1/y^3 -3/y +1 =0 . Multiply by y^3 both sides to get
1-3y^2+y^3 = 0 . This is the equation in y whose roots are 1/x1, 1/x2 ,1/x3.
Therefore the sum of the roots 1/x1+1/x2+1/x3 = -coefficient of y^2/ coefficient of y^3 = -(-3)/1 = 3