# Thread: proving help

1. ## proving help

The roots of the equation $x^2+px+1=0$ are $\alpha$ and $\beta$. If one of the roots of the equation $x^2+qx+1=0$ is $\alpha^3$ , prove that the other root is $\beta^3$ .

I tried this ..

For this equation $x^2+px+1=0$ ,

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

For the second equation
$x^2+qx+1=0$

$\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)$
$=(-p)[(-p)^2-2]$
$=-p^3+2p$

Obviously , i am wrong .. no idea where my mistake is , or i started wrongly .

2. $x^2+px+1 = (x-\alpha)(x-\beta) = x^2+x(-\beta-\alpha) + \beta \alpha$

Hence we can deduce that:

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

Now let's look at the 2nd equation!
$x^2+qx+1 = x^2+qx+\alpha \beta$

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

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

From this we can deduce that:

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

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

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

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

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

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

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

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

$y = \beta^3$

3. ## Re :

Lots thanks Mush , understooD !

4. Originally Posted by Mush
$x^2+px+1 = (x-\alpha)(x-\beta) = x^2+x(-\beta-\alpha) + \beta \alpha$

Hence we can deduce that:

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

Now let's look at the 2nd equation!
$x^2+qx+1 = x^2+qx+\alpha \beta$

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

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

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

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

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

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

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

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

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

$y = \beta^3$
should be
[tex]-\alpha^3-y[tex]

5. Originally Posted by repcvt
should be
[tex]-\alpha^3-y[tex]
True, that's a typo. Luckily it's irrelevant to the workings of the solution.

PS: html doesn't tend to work inside latex commands.