Find conditions on $a$ and $b$ given that two cubics are the same.

Question

The polynomial $x^3+x-3=0$ has roots $\alpha$, $\beta$ and $\gamma$. If $\frac{a\alpha+1}{\alpha-b}$, $\frac{a\beta+1}{\beta-b}$ and $\frac{a\gamma+1}{\gamma-b}$ are the roots of another cubic, what are the conditions on $a$ and $b$ given that the two cubics are the same?

Where should I start this? Using Vieta's formulas, this is tedious and I got stuck with too many unknowns. Please help

Answer 1

Hint:

Let $y=\dfrac{ax+1}{x-b}\implies x=\dfrac{1+by}{y-a}$

$$\implies\left(\dfrac{1+by}{y-a}\right)^3+\dfrac{1+by}{y-a}-3=0$$

Rearrange to form a cubic equation in $y$ like $$Ay^3+By^2+Cy+D=0$$

We need $$\dfrac 1A=\dfrac0 B=\dfrac1C=\dfrac{-3}D$$

Answer 2

Hint: If the two cubics are the same, then their roots are the same.