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?

When $f(x+1)=0$, we have,
$$(x+1)^3+a(x+1)^2+b(x+1)-2=0$$
Whose sum of roots can be obtained by "negative of coefficient of $x^2$ upon coefficient of $x^3$", which comes out to be $-\frac{3+a}{1}$.
Similarly, when $f(\frac 1x)=0$, we have,
$$\frac {1}{x^3}+a\frac{1}{x^2}+b\frac{1}{x}-2=0$$
Since $x\neq 0$, we can rearrange and get
$$-2x^3+bx^2+ax+1=0$$
Whose sum of roots is simply $\frac b2$. So,
$$\frac b2 = -3-a$$
Use that with the condition for $f(1)=5$ and solve for $a,b$. Then substitute $x=2$. Ta-da!
