If α, β and γ are the zeros of $x^3-7x^2+14x-8$, then 1/α + 1/β + 1/γ equals:

For $x^3+b{x}^2+cx+d$ the zeros satisfy α + β + γ = 7, αβ + βγ + γα = 14 and αβγ = 8. Then 1/α + 1/β + 1/γ = (αβ + βγ + γα)/(αβγ) = 14/8 = 7/4. (The zeros are 1, 2 and 4, and 1 + 1/2 + 1/4 = 7/4.)