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Given the roots of the cubic equation $x^3+4x^2+3x+2=0$ are $\alpha, \beta, \gamma$, determine the cubic equation with roots $\beta\gamma, \gamma\alpha, \alpha\beta$.

How on earth do I work out what the value of $\alpha '\beta '+\alpha '\gamma ' + \beta '\gamma '$?

I worked out that $\alpha '\ + \beta ' + \gamma ' = 3$.

Thanks :)

Avraham
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asa
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2 Answers2

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For typing speed, I hope you don't mind if I use latin letters :)

Let $a,b,c$ the roots. You know that $a+b+c=-4$, $ab+bc+ac=3$ and $abc=-2$. Let $a'=bc$, $b'=ac$, $c'=ab$.

Now you want to know:

  • $a'+b'+c'=ab+bc+ab=3$
  • $a'b'+b'c'+a'c'=abc^2+a^2bc+ab^2c=abc(a+b+c)=8$
  • $a'b'c'=a^2b^2c^2=(abc)^2= 4$

Therefore, your polynomial is $x^3-3x^2+8x-4$

ajotatxe
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For an alternative answer (and since @Community bumped the question to the main page): $\;\alpha\beta\gamma = -2$ by Vieta's, so the problem reduces to finding the polynomial with roots $\frac{-2}{\alpha}, \frac{-2}{\beta}, \frac{-2}{\gamma}\,$.

  • The polynomial with roots $y = \frac{1}{x}$ is $Q(y)=y^3P\left(\frac{1}{y}\right)=2y^3+3y^2+4y+1\,$.

  • The polynomial with roots $z = -2y$ is $R(z)=Q\left(\frac{-z}{2}\right)=\frac{-1}{4}\left(z^3-3z^2+8z-4\right)\,$.

Changing the variable back to $x\,$, the equation with roots $\frac{-2}{\alpha}, \frac{-2}{\beta}, \frac{-2}{\gamma}\,$ is $x^3-3x^2+8x-4=0$.

dxiv
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