Reciprocals of Roots in a Polynomial Equation

[Illmatic1]

Let $a, b, c, d, e$, and $f$ be the roots of $x^6 + 15x^5 + 53x^4 -127x^3 -1038x^2 -1832x- 960 = 0.$

Find $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f}.$

 

[anemone]

Let $a, b, c, d, e$, and $f$ be the roots of $x^6 + 15x^5 + 53x^4 -127x^3 -1038x^2 -1832x- 960 = 0.$

Find $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f}.$

My solution:

Spoiler

Note that we could easily create another polynomial with its roots the reciprocals of the roots of $x^6 + 15x^5 + 53x^4 -127x^3 -1038x^2 -1832x- 960 = 0$, i.e. $- 960x^6 -1832x^5 -1038x^4 -127x^3 +53x^2+15 x +1= 0$, therefore the sum of all its roots $=\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f}=-\frac{1832}{960}=-\frac{229}{120}$.