Polynomial transformations and Vieta's formulas 
Let $f(x)$ be a monic, cubic polynomial with $f(0)=-2$ and $f(1)=−5$. If the sum of all solutions to $f(x+1)=0$ and to $f\big(\frac1x\big)=0$ are the same, what is $f(2)$?

From $f(0)$ I got that $f(x)=x^3+ax^2+bx-2$ and from $f(1)=-5$ that $a+b = -4$ however I'm not sure how to use the info about the transformations to find $f(2).$ It seems that $(x+1)$ is a root for $f(x+1)$ and the same logic applies for $f\big(\frac1x\big)$?
Should I use Vieta's here or what's the appropriate way to go?

If $\alpha$ is a root of $f(x)$ then $\alpha - 1$ is a root of $f(x+1)$ (and vice-versa). So the sum of roots of $f(x+1)$ is the sum of roots of $f(x)$, minus $3$ which by Vieta' formula is $-a$ minus $3 =-a-3$.
Now, the roots of $f(\frac 1x) = \frac{-2x^3+bx^2+ax+1}{x^3}$  are also the roots of $-2x^3+bx^2+ax+1$, the reciprocal polynomial of $f$. The sum of roots of this polynomial is $\frac b2$ by Vieta's formula.
Finally, $-a-3= \frac b2$. Conclude.
