If $\alpha , \beta , \gamma$ are roots of equation, find values in terms of $p$ and $q$ Equation is $z^3-pz+q=0$
I have solved $\alpha^2+\beta^2+\gamma^2=2p$, $\alpha^3+\beta^3+\gamma^3=-3p$ and $\alpha^4+\beta^4+\gamma^4=2p^2$ already.
I am yet to solve the value of:
a. $(\alpha+\beta-\gamma)(\beta+\gamma-\alpha)(\gamma+\alpha-\beta)$
b. $\frac{1}{\alpha+\beta}+\frac{1}{\gamma+\beta}+\frac{1}{\alpha+\gamma}$.
c. $(\alpha\beta-1)(\beta\gamma-1)(\gamma\alpha-1)$
The equations scared me a bit and the constant cycle of expanding and factorising and subbing kept getting me incorrect answers. Is there a smarter way to get around these?
 A: Hints:
Note that $\alpha + \beta + \gamma = 0$ ; $\alpha\beta + \alpha\gamma + \beta\gamma = -p $ and $\alpha\beta\gamma = -q $.


*

*$(\alpha + \beta - \gamma)(\beta + \gamma - \alpha)(\gamma + \alpha - \beta) = (-2\alpha)(-2\beta)(-2\gamma) = ??$

*$\frac {1}{\alpha + \beta} + \frac {1}{\beta + \gamma} + \frac {1}{\alpha + \gamma} = -\frac {1}{\alpha} - \frac {1}{\beta} - \frac {1}{\gamma} = - (\frac {\alpha\beta + \beta\gamma + \alpha \gamma}{\alpha\beta\gamma}) = ??$

*$(\alpha\beta - 1)(\beta\gamma - 1)(\alpha \gamma - 1) = (\frac {-q}{\alpha}-1)(\frac {-q}{\beta} - 1)(\frac {-q}{\gamma} - 1) = ??$
A: Let $P(z)=z^3-pz+q$ with $\alpha, \beta, \gamma$ the three roots of $P(z)=0$
You should first show $\alpha+\beta+\gamma=0$ which gives you major simplifications.

Note also for powers, multiplying the original equation by $z^r$ gives $$z^{3+r}-pz^{1+r}+qz^r=0$$
Now substitute $\alpha, \beta, \gamma$ and add, and put $S_n=\alpha^n+\beta^n+\gamma^n$ to get $$S_{3+r}-pS_{1+r}+qS_r=0$$
(Note with care that $S_0=3$ and also $S_1=0$ as above). You can use this to check your results, some of which should include multiples of $q$. For example $S_3=-3q$, not $-3p$.

For the last one you could multiply through by $\gamma\alpha\beta$ using $\alpha\beta\gamma=-q$

$S_0=\alpha^0+\beta^0+\gamma^0=3$
$S_1=\alpha+\beta+\gamma=0$
$S_2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=0+2p=2p$
$S_3=pS_1-qS_0=-3q$
$S_4=pS_2-qS_1=2p^2$
Now note that you can write $2p=S_2=pS_0-qS_{-1}$ so that $S_{-1}=\frac pq$ and other sums of reciprocals of powers can be found by going down the recurrence rather than up it. Not so necessary for this example, but useful to note for the future.

For (c) note that $\alpha\beta\gamma =-q$ so that the expression becomes $$\frac {(\gamma+q)(\alpha+q)(\beta+q)}{q}$$
If we change the sign of $x$ in the original equation we obtain the equation for which $-\alpha,-\beta, -\gamma$ are the roots ie the polynomial $P(-z)$ factorises as $P(-z)=(z+\alpha)(z+\beta)(z+\gamma)$. Now what is $P(-q)$?
