Given a cubic equation $x^3+2x^2-5x+1=0$, find the equation with roots $\alpha\beta$, $\beta\gamma$ and $\gamma\alpha$. For the case where the new roots are ${\alpha}^2$, ${\beta}^2$ and ${\gamma}^2$, I can use the substitution $y=x^2$ but for this case I can't since the new roots are products of two old roots. However, for the case stated in the title above, I know this can be done where we set the equation to be $$x^3+px^2+qx+r$$ Then we can see that, $$-p=\Sigma{\alpha\beta}=-5\Rightarrow p=5$$ $$q=\Sigma{(\alpha\beta) (\beta\gamma) }$$ $$q=\alpha\beta\gamma\Sigma\alpha=(-1)(-2)\Rightarrow q=2$$ $$-r=(\alpha\beta) (\beta\gamma) (\gamma\alpha) $$ $$-r=(\alpha\beta\gamma)^2=(-1)^2\Rightarrow r=-1$$

However, this method is quite tedious, especially if the new roots are $\alpha+\beta$, $\beta+\gamma$ and $\gamma+\alpha$. So does substitution method exist for this condition or is the one shown above the only method?


2 Answers 2


For this particular equation we have $\alpha\beta\gamma=-1$ and so $$\alpha\beta=-\frac1\gamma\,,\ \beta\gamma=-\frac1\alpha\,,\ \gamma\alpha=-\frac1\beta\ .$$ So $\alpha\beta,\beta\gamma,\gamma\alpha$ are the roots of $$\Bigl(x+\frac1\alpha\Bigr)\Bigl(x+\frac1\beta\Bigr)\Bigl(x+\frac1\gamma\Bigr)$$ and also of $$\eqalign{\frac{\alpha\beta\gamma}{(-x)^3} \Bigl(x+\frac1\alpha\Bigr)\Bigl(x+\frac1\beta\Bigr)\Bigl(x+\frac1\gamma\Bigr) &=\Bigl(-\frac1x-\alpha\Bigr)\Bigl(-\frac1x-\beta\Bigr)\Bigl(-\frac1x-\gamma\Bigr)\cr &=f\Bigl(-\frac1x\Bigr)\ ,\cr}$$ where $f(x)$ is your original cubic.




As $\gamma$ is a root of the given equation,

replace the value of $\gamma$

For $x^2=y$

Square both sides of $$x(x^2-5)=-(2x^2+1)$$

Replace $x^2$ with $y$

  • $\begingroup$ So I substitute $y=-\frac{1}{x}$? $\endgroup$
    – T. Joel
    Jul 12, 2019 at 6:19
  • $\begingroup$ @T.Joel, yes, you are correct $\endgroup$ Jul 12, 2019 at 6:20
  • $\begingroup$ May I ask something similar? What about a cubic for roots ${\alpha}^2+\alpha$, ${\beta}^2+\beta$, ${\gamma}^2+\gamma$? It's a little difficult to find an expression for $x$ in terms of $y$ if I use $y=x^2+x$ $\endgroup$
    – T. Joel
    Jul 12, 2019 at 6:25
  • 1
    $\begingroup$ @Joel, $4y+1=(2x+1)^2$ First find the equation of $2x+1$ $\endgroup$ Jul 12, 2019 at 6:29

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