# Find a new cubic equation with new roots $\alpha\beta$, $\beta\gamma$ and $\gamma\alpha$. Can I use substitution for this?

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?

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.

$$y=\alpha\beta=\dfrac{-1}{\gamma}$$

$$\implies\gamma=?$$

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 $$y$$

• So I substitute $y=-\frac{1}{x}$? Jul 12, 2019 at 6:19
• @T.Joel, yes, you are correct Jul 12, 2019 at 6:20
• 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 $y=x^2+x$ Jul 12, 2019 at 6:25
• @Joel, $4y+1=(2x+1)^2$ First find the equation of $2x+1$ Jul 12, 2019 at 6:29