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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?

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Math goes in MathJax - the entire question body doesn't. I appreciate your apology, and I sincerely implore you to take a look here at how I've edited your question and try to emulate it in the future. Please see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. –  Zev Chonoles Jul 2 '13 at 9:24
    
The polynomial with roots $1/\alpha,1/\beta,1/\gamma$ is $dx^3-cx^2+bx-1$. –  Gerry Myerson Jul 2 '13 at 10:55
    
@Gerry Thanks for the hint. I tried this, but got to the same point as in my OP in a long and gruesome series of algebraic manipulations! I'm now checking a simple example to see if Lab's method works in practice. –  Guy Corrigall Jul 3 '13 at 2:08
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1 Answer

up vote 0 down vote accepted

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, $$(1 + \alpha^3)(1 + \beta^3)(1 + \gamma^3)=-D$$

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@Zev Many thanks, will read these references –  Guy Corrigall Jul 2 '13 at 9:42
    
@GuyCorrigall, observe that this method can be applied to other two cases as well –  lab bhattacharjee Jul 2 '13 at 9:46
    
Thank you. Your answer prompts me to transform the original polynomial into one with roots 1+α3 etc.I'll give this a go. –  Guy Corrigall Jul 2 '13 at 10:26
    
I made it! A very long and error prone series of algebraic manipulations, but confirmed correct by trialling a cubic with roots 1, 2 and 3 and substituting in your 'arranged' cubic whose coefficients depend only on b,c and d. One question, though; I can't quite see why simply cubing the expression for y in terms of alpha leads to the correct cubic with the correct roots (1+alpha, etc). Can you elucidate? –  Guy Corrigall Jul 3 '13 at 11:25
    
@GuyCorrigall, if you meant to cube $y=b\alpha^2-c\alpha+(1+d),$ it will tough to remove $\alpha$ from the expression. Also observe that here we only need the coefficient of $y^3$ which is $1$ here, the constant term $(D)$ which is $-(1+d)^3-b^3+c^3+3bc(1+d)$ –  lab bhattacharjee Jul 3 '13 at 11:58
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