Sorry for formatting issues as I’m on my phone.
The question is: the cubic equation $ax^3+bx^2+cx+d=0$ has roots $α$ , $β$, and $γ$. Find a cubic equation that has roots $1/α$ , $1/β$ and $1/γ$.
The only way I can see how to do it using Vietas formula is doing 1/-b/a , 1/c/a and so forth, but I don’t think that’s the correct way to do it so I’d like some guidance.
For $d\neq0$ it's just $$a\left(\frac{1}{x}\right)^3+b\left(\frac{1}{x}\right)^2+c\left(\frac{1}{x}\right)+d=0$$ or $$dx^3+cx^2+bx+a=0.$$ Indeed, since $\alpha$ is root of the equation $ax^3+bx^2+cx+d=0$, so $$a\alpha^3+b\alpha^2+c\alpha+d=0$$ or $$a\left(\frac{1}{\frac{1}{\alpha}}\right)^3+b\left(\frac{1}{\frac{1}{\alpha}}\right)^2+c\left(\frac{1}{\frac{1}{\alpha}}\right)+d=0,$$ which says that $\frac{1}{\alpha}$ is a root of the equation $$a\left(\frac{1}{x}\right)^3+b\left(\frac{1}{x}\right)^2+c\left(\frac{1}{x}\right)+d=0.$$ Similarly, for $\frac{1}{\beta}$ and for $\frac{1}{\gamma}.$
By the Viete's theorem we obtain the same result: $$\alpha+\beta+\gamma=-\frac{b}{a},$$ $$\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}$$ and $$\alpha\beta\gamma=-\frac{d}{a}.$$ Thus, $$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\alpha\beta+\alpha\gamma+\beta\gamma}{\alpha\beta\gamma}=\frac{\frac{c}{a}}{-\frac{d}{a}}=-\frac{c}{d},$$ $$\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\gamma\alpha}=\frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}=\frac{-\frac{b}{a}}{-\frac{d}{a}}=\frac{b}{d}$$ and $$\frac{1}{\alpha\beta\gamma}=\frac{1}{-\frac{d}{a}}=-\frac{a}{d}.$$ Can you end it now?