Relationship between coefficients of a polynomials and its roots. The roots of the equation $x^4 -3x^2 + 5x - 2 = 0$ are $\alpha$, $\beta$, $\gamma$ and $\delta$.   $\alpha^n + \beta^n + \gamma^n + \delta^n$ is denoted by $S(n)$. Find values of $S(2)$ and $S(4)$ and of $S(3)$ and $S(5)$. 
Hence, find the value of 
$\alpha^2 (\beta^3 + \gamma^3 + \delta^3) + \beta^2 ( \alpha^3 + \gamma^3 + \delta^3) + \gamma^2 (\alpha^3 + \beta^3 + \delta^3) + \delta^2 (\alpha^3 + \beta^3 + \gamma^3)$
This topic is a rather strange one, and I can't seem to locate the formulas I need, and I need them for my exam. If you do use any formula (except $\sum\alpha$, $\sum\alpha\beta$, $\sum\alpha\beta\gamma$ and $\alpha\beta\gamma\delta$), please state it. Thank you in advance.
 A: $$S(2)=\left(\sum\alpha\right)^2-2\sum\alpha\beta$$
$$S(3)=\left(S(2)-\sum\alpha\beta\right)\left(\sum\alpha\right)+3\sum\alpha\beta\gamma$$
See here to verify the second identity.Now, summing cyclically we have:
$$S(4)-3S(2)+5S(1)-8=0$$
More generally, multiplying by $x^n$ and summing cyclicaclly we get the recurrence relation:
$$S(n+4)-3S(n+2)+5S(n+1)-2S(n)=0$$

Finally, for your bizarre looking last expression, we have
$$\sum a^2\left(S(3)-\alpha^3\right)=S(3)\sum a^2-\sum\alpha^5=S(2)S(3)-S(5)$$
A: Denote 
$A(1) = \alpha+\beta+\gamma+\delta  = 0$;

$A(2) = \alpha\beta+...+\gamma\delta  = -3$;

$A(3) = \alpha\beta\gamma +...+\beta\gamma\delta  = -5$;

$A(4) = \alpha\beta\gamma\delta  = -2$.
Applying Newton's identities, we obtain
$S(1) = A(1) = 0$;

$S(2) = A(1) S(1) - 2A(2) = 0+6=6$;

$S(3) = A(1) S(2) - A(2)S(1) + 3A(3) = 0+0-15=-15$;

$S(4) = A(1) S(3) - A(2)S(2) + A(3)S(1) - 4A(4) = 0+18+0+8=26$;

$S(5) = A(1) S(4) - A(2)S(3) + A(3)S(2) - A(4)S(1) + 5A(5) = 0-45-30+0+0=-75$

(one can obtain last identity, when consider equation $x^5-3x^3+5x^2-2x=0$ with $5$ roots $\alpha,\beta,\gamma,\delta,0$, where $A(5)=0$, and previous $A(n)$ are the same).

Easy to see, that 
$$\alpha^2 (\beta^3 + \gamma^3 + \delta^3) + \beta^2 ( \alpha^3 + \gamma^3 + \delta^3) + \gamma^2 (\alpha^3 + \beta^3 + \delta^3) + \delta^2 (\alpha^3 + \beta^3 + \gamma^3)$$
$$=\alpha^2 (\alpha^3+\beta^3 + \gamma^3 + \delta^3) + \beta^2 ( \alpha^3 + \beta^3+\gamma^3 + \delta^3) + \gamma^2 (\alpha^3 + \beta^3 + \gamma^3+\delta^3) + \delta^2 (\alpha^3 + \beta^3 + \gamma^3 + \delta^3) - \alpha^5-\beta^5-\gamma^5-\delta^5$$
$$
= (\alpha^2+\beta^2 + \gamma^2 + \delta^2)(\alpha^3+\beta^3 + \gamma^3 + \delta^3)
- \alpha^5-\beta^5-\gamma^5-\delta^5 = S(2)S(3)-S(5).
$$
$$
= 6\cdot (-15) - (-75) = -90+75=-15.
$$

Note:
equation $x^4-3x^2+5x-2=0$ has $2$ real and $2$ complex roots:

$x_1\approx -2.344470$;

$x_2\approx 0.578277$;

$x_{3,4} \approx 0.883096 \pm 0.833866i$.
