<s>[INST] Relation between roots and coefficients of equation If the roots of the equation $x^4 - x^3 +2x^2+x+1 = 0 $ are given by $a,b,c,d$ then find the value of $(1+a^3)(1+b^3)(1+c^3)(1+d^3)$
I found out that:
$$ (1+a^3)(1+b^3)(1+c^3)(1+d^3) = (abcd)^3+\sum(abc)^3 +\sum(ab)^3+\sum(a)^3 +1$$
But how do I calculate $\sum(abc)^3$ and $\sum(ab)^3$?
 [/INST] Something seemed a bit suspicious about that polynomial:  poking around a little, I found that
$$ ( x^4 \ - \ x^3 \ + \ 2x^2 \ + \ x \ + \ 1)·(x^2 \ + \ x \ - \ 1 ) \ \ = \ \ x^6 \ + \ 4x^3 \ - \ 1 \ \ . $$
Treating this as a "quadratic in $ \ x^3 \ \ , $ " we find that there are two "triplets" of zeroes given by $ \ x^3 \ = \ -2 \ \pm \ \sqrt5 \ \ . $  The factor $ \ (x^2 \ + \ x \ - \ 1 ) \ $ has the "familiar" real zeroes $ \ -\frac12 \ + \ \frac{\sqrt{5}}{2} \ = \ \frac{1}{\phi} \ \   $ and $ \ -\frac12 \ - \ \frac{\sqrt{5}}{2} \ = \ -\phi \ \   $ (the Golden Ratio emerges from the shadows once more!).  The cubes of these numbers are indeed $ \ \frac{1}{\phi^3} \ = \ -2 + \sqrt5 \  $  and  $ \ (-\phi)^3  \ = \ -2 - \sqrt5 \ \ ,  $  so the other four zeros of $ \  x^6 \ + \ 4x^3 \ - \ 1 \ \ , $ and thus the four zeroes of $ \ x^4 \ - \ x^3 \ + \ 2x^2 \ + \ x \ + \ 1 \ \ , $ are the four complex(-conjugate) cube-roots of $  \ -2 + \sqrt5 \  $  and  $  \ -2 - \sqrt5 \ \ . $
Consequently, we may designate $ \ a^3 \ , \ b^3 \ = \ -2 \ + \ \sqrt5 \ \ , $ with $ \ b \ = \ \overline{a} \ \ , $ and $ \ c^3 \ , \ d^3 \ = \ -2 \ - \ \sqrt5 \ \ , $ with $ \ d \ = \ \overline{c} \ \ . $  We then seek the value of
$$  (1 + a^3)·(1 + b^3)·(1 + c^3)·(1 + d^3) \ \ = \ \ (1 + a^3)·(1 + \overline{a}^3)·(1 + c^3)·(1 + \overline{c}^3)   $$
$$ = \ \ ( \ 1 \ +  \ a^3 \ + \ \overline{a}^3 \ + \ [a·\overline{a}]^3 \ )·( \ 1 \ +  \ c^3 \ + \ \overline{c}^3 \ + \ [c·\overline{c}]^3 \ ) \ \ .   $$
It remains to determine the last terms in each of the factors above.  Since $ \ a \ , \ \overline{a} \ , $ and $ \ \alpha \ = \ -\frac12 \ + \ \frac{\sqrt{5}}{2} \ $ all have the cube $  \ -2 + \sqrt5 \ , $ it follows that $ \ (a·\overline{a}·\alpha)^3 \ = \  (-2 + \sqrt5)^3 \ \ , $ and hence $ \ (a·\overline{a})^3 \ = \  (-2 + \sqrt5)^2 $ $ \ = \ 9 \ - \ 4\sqrt5 \ \ ; \  $ similarly, $ \ (c·\overline{c})^3 \ = \  (-2 - \sqrt5)^2 $ $ \ = \ 9 \ + \ 4\sqrt5 \ \ .  $
So at last we obtain
$$  (1 + a^3)·(1 + b^3)·(1 + c^3)·(1 + d^3) $$ $$ =  \ \ ( \ 1 \ +  \ [-2 + \sqrt5] \ + \ [-2 + \sqrt5] \ + \ [9 - 4\sqrt5] \ )$$ $$· \ ( \ 1 \ +  \ [-2 - \sqrt5] \ + \ [-2 - \sqrt5] \ + \ [9 + 4\sqrt5] \ )     $$
$$ =  \ \ ( \ 6 \ -  \ 2\sqrt5 \ ) \ · \ ( \ 6 \ +  \ 2\sqrt5 \ ) \ \ = \ \ 36 \ - \ 20 \ \ = \ \ 16 \ \ .    $$
[Evidently, this approach has connections to the more sophisticated methods employed by dxiv and Snacc.]
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