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**Mush** $\displaystyle x^2+px+1 = (x-\alpha)(x-\beta) = x^2+x(-\beta-\alpha) + \beta \alpha $

Hence we can deduce that:

$\displaystyle \alpha \beta = 1 $

$\displaystyle \therefore \alpha = \frac{1}{\beta} $

Now let's look at the 2nd equation!

$\displaystyle x^2+qx+1 = x^2+qx+\alpha \beta $

But we also know that $\displaystyle \alpha^3$ is a root (and hence $\displaystyle (x-\alpha^3) $ is a factor), and we need to find the other root, call it $\displaystyle y$ (hence $\displaystyle (x-y)$ is a factor!) Hence:

$\displaystyle x^2+qx+\alpha \beta = (x-\alpha^3)(x-y) = x^2+x(\alpha^3-y) + \alpha^3y$

should be -\alpha^3-y

From this we can deduce that:

Hence $\displaystyle \alpha^3y = \alpha \beta $

And with a bit of manipulation, and then plugging in the result from the first equation that $\displaystyle \alpha = \frac{1}{\beta}$, then we can find $\displaystyle y$

$\displaystyle y = \frac{\alpha \beta}{\alpha^3} $

$\displaystyle y = \frac{ \beta}{\alpha^2} $

$\displaystyle y = \frac{\beta}{(\frac{1}{\beta})^2} $

$\displaystyle y = \frac{\beta}{\frac{1}{\beta^2}} $

$\displaystyle y = \beta \times (\frac{1}{\beta^2})^{-1}$

$\displaystyle y = \beta \times \beta^2$

$\displaystyle y = \beta^3 $