Vieta's formula for $f(\frac{1}{x})$ 
Let $f(x)$ be a monic cubic polynomial. Let the solutions of the equation $f(\frac{1}{x})=0$ be $\alpha, \beta$ and $\gamma$. If $\alpha+\beta+\gamma = 10$ and $\alpha\beta\gamma=15$, what is $\left \lceil{f(10)-f(-10)}\right \rceil$?

So since $f(x)$ is a monic cubic we can write it as $f(x) = x^3+bx^2+cx+d$ and from here $f(\frac{1}{x}) = \frac{1}{x^3} + b \frac{1}{x^2} + c \frac{1}{x}+d$.
The question stated that $\alpha, \beta$ and $\gamma$ were solutions for $f(\frac{1}{x})$ and I see that this would imply to use Vieta's here, but how do we go about it with something like $f(\frac{1}{x})$?
 A: First notice that:
\begin{gather}
f(10)-f(-10) = 10^3+10^2b+10c+d - (-10)^3 - (-10)^2b-(-10)c -d =\\
=2000+20c
\end{gather}
So if we find the value of $c$ we are done.
As you noticed $f\left(\frac{1}{x}\right) = \frac{1}{x^3}(dx^3+cx^2+bx+1) = \frac{1}{dx^3}\left(x^3+\frac{c}{d}x^2+\frac{b}{d}x+\frac{1}{d}\right)$ where $d$ is different to zero because by hypotesis we have three solutions.
Thanks to Vieta's formula we have 
\begin{gather}
\frac{c}{d} = -(\alpha + \beta + \gamma) =-10\\
\frac{1}{d} =  -(\alpha\beta\gamma) -15
\end{gather}
Hence $c=\frac{2}{3}$ and the result is:
$$
f(10)-f(-10) =2000+20c = \frac{6040}{3}
$$
A: Hint. You only need to find the coefficients of $f(x),$ then you would easily continue. To do this, note that the equation $f(1/x)=0$ simplifies to $$1+bx+cx^2+dx^3=0,$$ whence $$x^3+\frac cdx^2+\frac bdx+\frac 1d=0,$$ and we see that $$\alpha\beta\gamma=-\frac 1d=15,$$ from where $d=-1/15.$
Also, we see that $$\alpha+\beta+\gamma=-\frac cd=10\implies c=150.$$
Finally, note that you really don't need to know the coefficients of the even terms of $f(x)$ since they will vanish in a substitution of the form $f(x)-f(-x).$
