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

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

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

• Doesn't it mean that $A=C=1, B=0, D=-3$? Could you explain $\dfrac 1A=\dfrac0 B=\dfrac1C=\dfrac{-3}D$? Commented Sep 10, 2019 at 8:13
• @T.Joel, $$x^3+x-2=0\iff 3x^3+3x-6=0$$ right? Commented Sep 10, 2019 at 8:16
• Also, isn't it $x=\frac{1+by}{y-a}$ from $y=\frac{ax+1}{x-b}$? Commented Sep 10, 2019 at 8:17
• @T.Joel, Sorry for the typo. coefficients will have same ratio Commented Sep 10, 2019 at 8:56
• @T.Joel, We can safely ignore the nature or the exact values of the roots in the current context Commented Sep 10, 2019 at 9:27

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

• So $\alpha=\frac{a\alpha+1}{\alpha-b}$? Commented Sep 10, 2019 at 8:11