<s>[INST] 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.
 [/INST] 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.
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