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`** .

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