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If $\alpha,\beta,\gamma$ are the roots of $x^3+qx+r=0(r\ne0)$ then find the equation whose roots are $\frac{\alpha}{\beta},\frac{\alpha}{\gamma},\frac{\beta}{\gamma},\frac{\beta}{\alpha},\frac{\gamma}{\alpha},\frac{\gamma}{\beta}.$

My book has solved it as follows:

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But I don't understand how can I be sure that the last equation is necessarily the required one.

I could not find any logical justification behind the claim that the last equation has the required roots. Is the solution correct? How? If not than What should be the correct solution?

Please help.

Jave
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  • I am not clear what you don't follow. The book derives $\beta=-r(y^2+y+1)/(qy(y+1))$. Cubing: $\beta^3=-r^3(y^2+y+1)^3/(qy(y+1))^3$. Then equate that to the $\beta^3=r/(y(y+1))$ expression (which comes from two lines further up) and you get the final equation. – almagest Sep 29 '19 at 16:43
  • Or are you worried that it has only been proved for $\alpha/\beta$, not for the other five possibilities? Well, an identical argument works for them too (and note that the equation has degree 6, so it has 6 roots). The only point that requires a little thought is why the book argument doesn't prove that $y=\alpha/\alpha=1$ is a root of the final equation. – almagest Sep 29 '19 at 16:50

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