Transformation of roots of a polynomial Suppose there is a cubic polynomial in x with roots A,B,C and another cubic polynomial (in t) with roots
1/(A-1),     1/(B-1), 1/(C-1)
is to be found.
My text book mentions two ways of doing this. One is to use vieta's relations, which is time consuming.
Another way mentioned is to let t=1/(x-1), make x the subject, and then substitute and simplify. However, I cant really wrap my mind around this. Just because this is the relation between the roots doesnt mean that all values of x and t are in thay relation, does it?
Moreover, if I try this method with more generality i.e trying to find a polynomial with roots f(A),f(B),f(C) where f(u) is some invertible function like e^u, it appears to fail.
Could someone please elaborate on the proper method of tackling such questions.
 A: Suppose you had the cubic polynomial $P(x) = ax^3 + bx^2 + cx + d = 0$ for which you had three real or complex roots $x_1, x_2, x_3$, then you can say that $P(x_i) = ax_i^3 + bx_i^2 + cx_i + d = 0$ for $i = 1,2,3$.
Now say you wanted to construct a new polynomial equation that had roots $y_1, y_2, y_3$ where $y_i = \frac 1{x_i-1}$ (contingent on the assumption none of the roots of the original is $1$, of course), then as your reference said, you can change the subject to $x_i$, i.e. $x_i = 1 + \frac 1{y_i}$ and substitute that into the original equation. You know that $P(x_i) = 0$, so that implies that $P(1+\frac 1{y_i}) = 0$ also. By manipulation, you can show that this new equation is a cubic $Q(y_i) = 0$. By the explicit mapping that you've already derived, each of the three roots $y_i$ of this new cubic will correspond to each of the three roots $x_i$ of the original cubic.
Certainly, this method will work for any invertible transform as long as you are careful about domain and existence considerations, but you may not get a polynomial equation after transformation (like in your case with the exponential function).
A: Let $A$, $B$ and $C$ be roots of the polynomial $x^3+ax^2+bx+c.$
Thus, $$A+B+C=-a,$$ $$AB+AC+BC=b,$$ $$ABC=-c$$ and we obtain:
$$\sum_{cyc}\frac{1}{A-1}=\frac{\sum\limits_{cyc}(A-1)(B-1)}{\prod\limits_{cyc}(A-1)}=\frac{b+2a+3}{-c-b-a-1},$$ which gives a coefficient before $x^2$ in new polynomial:
$$\frac{2a+b+3}{a+b+c+1}.$$
Also, $$\sum_{cyc}\frac{1}{(A-1)(B-1)}=\frac{\sum\limits_{cyc}(A-1)}{\prod\limits_{cyc}(A-1)}=\frac{-a-3}{-a-b-c-1}=\frac{a+3}{a+b+c+1}$$ and
$$\prod_{cyc}\frac{1}{A-1}=\frac{1}{-a-b-c-1}$$ and we got the answer:
$$x^3+\frac{2a+b+3}{a+b+c+1}x^2+\frac{a+3}{a+b+c+1}x+\frac{1}{a+b+c+1}.$$
We used the Viete's theorem and this is a right method for solving your problem.
Also, you can see that by using a cyclic summation it turns out easy enough.
I think the second method takes more time.
Let $x=\frac{1}{y-1}$, where $y$ is a root if the polynomial $x^3+ax^2+bx+c.$
Thus, since $y=\frac{x+1}{x},$ we obtain:
$$\left(\frac{x+1}{x}\right)^3+a\left(\frac{x+1}{x}\right)^2+b\left(\frac{x+1}{x}\right)+c,$$ which gives the same result, but with a bit of more computations.
