Stuck on a further pure question, asked one of my maths teachers today but she hadn't done most of the paper in 20 years.

Here's the question:

Given that the roots of the cubic equation

$\displaystyle x^3+4x^2+3x+2=0$

are $\displaystyle \alpha, \beta, \gamma$ determine the cubic equation with roots $\displaystyle \beta \gamma , \gamma \alpha , \alpha \beta$

All I have so far is

$\displaystyle \alpha + \beta + \gamma = -4 \: ; \: \beta \gamma + \gamma \alpha + \alpha \beta = 3 \: ; \: \alpha \beta \gamma = -2$

I assume the new equations will looks like:

$\displaystyle x^3-(\beta \gamma , \gamma \alpha , \alpha \beta)x^2+(\alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma ^2)x - \alpha^2 \beta^2 \gamma^2$

I'm pretty sure the coefficient of the $\displaystyle x^2$ term will be $\displaystyle \beta \gamma + \gamma \alpha + \alpha \beta$ which equals 3.

And for the constants:

$\displaystyle \alpha \beta \gamma = -2$

$\displaystyle (\alpha \beta \gamma)^2 = 4$

$\displaystyle \alpha^2 \beta^2 \gamma^2 - 2(\beta \gamma + \gamma \alpha + \alpha \beta)=4$

$\displaystyle \alpha^2 \beta^2 \gamma^2 -2 \times 3 = 4$

$\displaystyle \alpha^2 \beta^2 \gamma^2 = 10$

I think I'm correct so far (may be some typos though), but I don't know how to get the x term.

Thanks.

edit

I got it, just factoring the coefficients of the x term:

$\displaystyle \alpha \beta \gamma (\alpha + \beta + \gamma) $

brain ache now.