If $\alpha,\beta,\gamma$ are the roots of $x^3+x^2-x+1=0$, find the value of $\prod\left(\frac1{\alpha^3}+\frac1{\beta^3}-\frac1{\gamma^3}\right)$

Question

If $\alpha,\beta,\gamma$ are the roots of the equation $x^3+x^2-x+1=0$, find the value of $\prod\left(\frac1{\alpha^3}+\frac1{\beta^3}-\frac1{\gamma^3}\right)$

My Attempt:

$\alpha+\beta+\gamma=-1$

$\alpha\beta+\beta\gamma+\gamma\alpha=-1$

$\alpha\beta\gamma=-1$

$\prod\left(\frac1{\alpha^3}+\frac1{\beta^3}-\frac1{\gamma^3}\right)=\prod\left(\frac{(\beta\gamma)^3+(\alpha\gamma)^3-(\alpha\beta)^3}{(\alpha\beta\gamma)^3}\right)=-\prod\left((\beta\gamma)^3+(\alpha\gamma)^3-(\alpha\beta)^3\right)$

Now, one option is to expand the product and put the values where required, but is there a shorter, better approach to do this?

Answer 1

1. Let $P(x) = x^3 + x^2 - x + 1$. Since all the zeros of $P(x)$ are non-zero, $\gcd(P(x), x^3) = 1$. Then the Bézout's identity tells that we can find polynomials $A(x)$ and $B(x)$ satisfying

$$ A(x) P(x) + B(x) x^3 = 1. $$

Although there is a systematic way of determining $A(x)$ and $B(x)$, called the extended GCD algorithm, it is not hard to see that $A(x) = x+1$ and $B(x) = -x-2$ from the computation

$$ (x+1)(x^2 - x + 1) = x^3 + 1 \qquad\implies\qquad (x+1)P(x) = x^4 + 2x^3 + 1. $$

The upshot of this computation is that, if $x = x_0$ is any zero of $P(x) = 0$, then

$$ B(x_0) x_0^3 = 1 \qquad\text{and hence}\qquad \frac{1}{x_0^3} = B(x_0) = -x_0-2. $$

2. Plugging this to OP's product, we get

$$ \prod_{\text{cyc}} \left( \frac{1}{\alpha^3} + \frac{1}{\beta^3} - \frac{1}{\gamma^3} \right) = \prod_{\text{cyc}} \left( -\alpha -\beta + \gamma - 2 \right) = \prod_{\text{cyc}} \left(2\gamma - 1\right) = (-2)^3 P\left(\frac{1}{2}\right) = -7. $$

Answer 2

Here is the method with least calculation I can think of.

I write $a,b,c$ for $\alpha,\beta,\gamma$ instead.

Let $p=a+b+c=-1$, $q=ab+bc+ca=-1$, $r=abc=-1$. Thus

$$a^3+b^3+c^3=p^3-3pq+3r=-7$$

Let $p'=1/a+1/b+1/c=q/r=1$, $q'=1/ab+1/bc+1/ca=p/r=1$, $r'=1/abc=-1$. Thus

$$a^{-3}+b^{-3}+c^{-3}=p'^3-3p'q'+3r'=-5$$

Thus we have

$$a^{-3}b^{-3}+b^{-3}c^{-3}+c^{-3}a^{-3}=(a^3+b^3+c^3)(a^{-3}b^{-3}c^{-3})=-7\times -1=7$$

By Vieta's theorem, $a^{-3}, b^{-3}, c^{-3}$ are roots of equation $x^3+5x^2+7x+1=0$.

So we have $a^{-3}+b^{-3}-c^{-3}=-5-2c^{-3}$ are roots of equations $(\frac{-5-t}{2})^3+5(\frac{-5-t}{2})^2+7(\frac{-5-t}{2})+1=-\frac{1}{8}x^3+\dots-\frac{7}{8}$ (the middle two terms no need to calculate.) So, we have $\prod (a^{-3}+b^{-3}-c^{-3})=-7$

Answer 3

Let $s=\frac1{\alpha^3}+\frac1{\beta^3}+\frac1{\gamma^3}$. Then the product we are looking for is $$ p(s)=\left(s-\frac2{\alpha^3}\right)\left(s-\frac2{\beta^3}\right)\left(s-\frac2{\gamma^3}\right)\tag1 $$ where $p$ is the monic polynomial whose roots are $\frac2{\alpha^3},\frac2{\beta^3},\frac2{\gamma^3}$.

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If $\alpha,\beta,\gamma$ are the roots of $$ x^3+x^2-x+1\tag2 $$ then $\frac1\alpha,\frac1\beta,\frac1\gamma$ are the roots of $$ x^3-x^2+x+1\tag3 $$ Reading off the coefficients of $(3)$, we get $$ \begin{align} \frac1\alpha+\frac1\beta+\frac1\gamma&=1\tag{4a}\\ \frac1{\alpha\beta}+\frac1{\beta\gamma}+\frac1{\gamma\alpha}&=1\tag{4b}\\ \frac1{\alpha\beta\gamma}&=-1\tag{4c}\\ \end{align} $$ Applying $\text{(4a)}$ and $\text{(4b)}$, we get $$ \begin{align} \frac1{\alpha^2}+\frac1{\beta^2}+\frac1{\gamma^2} &=\left(\frac1\alpha+\frac1\beta+\frac1\gamma\right)^2-2\left(\frac1{\alpha\beta}+\frac1{\beta\gamma}+\frac1{\gamma\alpha}\right)\tag{5a}\\ &=-1\vphantom{\frac12}\tag{5b} \end{align} $$ Applying $(3)$, then $\text{(4a)}$ and $(5)$ we get $$ \begin{align} \overbrace{\frac1{\alpha^3}+\frac1{\beta^3}+\frac1{\gamma^3}}^s &=\overbrace{\frac1{\alpha^2}-\frac1{\alpha}-1}^{1/\alpha^3}+\overbrace{\frac1{\beta^2}-\frac1{\beta}-1}^{1/\beta^3}+\overbrace{\frac1{\gamma^2}-\frac1{\gamma}-1}^{1/\gamma^3}\tag{6a}\\ &=-5\tag{6b}\vphantom{\frac12} \end{align} $$ That is, $s=-5$.

Applying $\text{(4c)}$, we get $$ \frac1{\alpha^3\beta^3\gamma^3}=-1\tag7 $$ Since $x^6\equiv-3x^2+2x+2$ mod $\left(x^3-x^2+x+1\right)$, we can apply $\text{(4a)}$ and $(5)$ to get $$ \begin{align} \frac1{\alpha^6}+\frac1{\beta^6}+\frac1{\gamma^6} &=\overbrace{-\frac3{\alpha^2}+\frac2{\alpha}+2}^{1/\alpha^6}\overbrace{\,-\,\frac3{\beta^2}+\frac2{\beta}+2}^{1/\beta^6}\overbrace{\,-\,\frac3{\gamma^2}+\frac2{\gamma}+2}^{1/\gamma^6}\tag{8a}\\ &=11\tag{8b}\vphantom{\frac12} \end{align} $$ Putting together $(6)$ and $(8)$, we get $$ \begin{align} \frac1{\alpha^3\beta^3}+\frac1{\beta^3\gamma^3}+\frac1{\gamma^3\alpha^3} &=\frac12\left[\left(\frac1{\alpha^3}+\frac1{\beta^3}+\frac1{\gamma^3}\right)^2-\left(\frac1{\alpha^6}+\frac1{\beta^6}+\frac1{\gamma^6}\right)\right]\tag{9a}\\ &=7\tag{9b}\vphantom{\frac12} \end{align} $$ Thus, using $(6)$, $(7)$, and $(9)$, $\frac1{\alpha^3},\frac1{\beta^3},\frac1{\gamma^3}$ are roots of $$ x^3+5x^2+7x+1\tag{10} $$ and $\frac2{\alpha^3},\frac2{\beta^3},\frac2{\gamma^3}$ are roots of $$ p(x)=x^3+10x^2+28x+8\tag{11} $$

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Using $(6)$ and $(11)$, the product we are looking for is $$ p(s)=-7\tag{12} $$