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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

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

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  • $\begingroup$ Doesn't it mean that $A=C=1, B=0, D=-3$? Could you explain $\dfrac 1A=\dfrac0 B=\dfrac1C=\dfrac{-3}D$? $\endgroup$ – T. Joel Sep 10 at 8:13
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    $\begingroup$ @T.Joel, $$x^3+x-2=0\iff 3x^3+3x-6=0$$ right? $\endgroup$ – lab bhattacharjee Sep 10 at 8:16
  • $\begingroup$ Also, isn't it $x=\frac{1+by}{y-a}$ from $y=\frac{ax+1}{x-b}$? $\endgroup$ – T. Joel Sep 10 at 8:17
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    $\begingroup$ @T.Joel, Sorry for the typo. coefficients will have same ratio $\endgroup$ – lab bhattacharjee Sep 10 at 8:56
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    $\begingroup$ @T.Joel, We can safely ignore the nature or the exact values of the roots in the current context $\endgroup$ – lab bhattacharjee Sep 10 at 9:27
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Hint: If the two cubics are the same, then their roots are the same.

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  • $\begingroup$ So $\alpha=\frac{a\alpha+1}{\alpha-b}$? $\endgroup$ – T. Joel Sep 10 at 8:11

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