<s>[INST] Expressing polynomial roots expression in terms of coefficients This is my first question on MSE. Apologies in advance for any textual or LaTeX errors.
I'm stuck with this problem:

Given  $x^3 - bx^2 + cx - d = 0$ has roots $\alpha$, $\beta$, $\gamma$, find an expression in terms of $b$, $c$ and $d$ for:
  (i) $\alpha^2 + \beta^2 + \gamma^2$
  (ii) $\alpha^3 + \beta^3 + \gamma^3$
  (iii) $(1 + \alpha^3)(1 + \beta^3)(1 + \gamma^3)$

I had no trouble with (i) or (ii), but got stuck on (iii) as follows: Expanding,
$$\begin{align*}
(1 + \alpha^3)(1 + \beta^3)(1 + \gamma^3) & = (1 + \alpha^3 + \beta^3 + \alpha^3\beta^3)(1 + \gamma^3)\\ 
& = 1 + (\alpha^3 + \beta^3 + \gamma^3) + (\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3) + \alpha^3\beta^3\gamma^3
\end{align*}$$
The first, second and fourth RHS terms are no problem, leaving us with:
$$\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3 = \left(\frac{1}{\gamma^3} + \frac{1}{\alpha^3} + \frac{1}{\beta^3} \right)\alpha^3\beta^3\gamma^3$$
So now we are left with the term in brackets. My next thought was to transform the original polynomial to one with roots $\frac{1}{\alpha}$, $\frac{1}{\beta}$ and $\frac{1}{\gamma}$ and then use the answer to (ii) above. Will this work? Or is there a better approach?
 [/INST] Let $y=1+\alpha^3$
Again, $\alpha^3=b\alpha^2-c\alpha+d$
$$\implies y-1-d=b\alpha^2-c\alpha$$
Cubing  we get  $$ (y-1-d)^3=b^3\alpha^6-c^3\alpha^3-3bc\alpha^2\cdot \alpha(b\alpha^2-c\alpha)$$
$$(y-1-d)^3=b^3(y-1)^2-c^3(y-1)-3bc(y-1)(y-1-d)$$ as $\alpha^3=y-1,b\alpha^2-c\alpha=y-1-d$
Arrange as $y^3+By^2+Cy+D=0$ whose roots are $1 + \alpha^3,1 + \beta^3,1 + \gamma^3$
Using Vieta's formulas,
$$(1 + \alpha^3)(1 + \beta^3)(1 + \gamma^3)=-D$$
</s>