1. ## proving help

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

I tried this ..

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

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

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

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

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

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

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:

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

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 $\displaystyle y$

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

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

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

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

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

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

$\displaystyle y = \beta^3$

3. ## Re :

Lots thanks Mush , understooD !

4. Originally Posted by 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 $\displaystyle y$

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

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

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

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

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

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

$\displaystyle y = \beta^3$
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.