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 $\sum(ab)^3$?
 A: With a lot of these problems involving symmetric sums of the roots of a polynomial, there are two ways to do it. There is a long and tedious way using Vieta’s formulas and Newton sums, and a “clever” way that just involves evaluating the polynomial a couple times. I will go over the clever way for this problem, and I’ll encourage you to look into vieta’s formulas and Newton sums yourself.
For the clever way, we start by factoring the polynomial:
$$P(x)=x^4-x^3+2x^2+x+1=(x-a)(x-b)(x-c)(x-d)$$
Now, observe that if we let $x=-1$, something interesting happens:
$$P(-1)=(-1-a)(-1-b)(-1-c)(-1-d)=(1+a)(1+b)(1+c)(1+d)$$
This is related to the product we wish to evaluate, but it doesn’t have the exponents. We can fix this by noticing that our product can be factored in the complex numbers. If we let $\omega=e^{\frac{2\pi i}{3}}$, then we have:
$$(1+a^3)(1+b^3)(1+c^3)(1+d^3)=(1+a)(1+\omega a)(1+\omega^2 a)(1+b)(1+\omega b)\ldots$$
$$=(-1-a)(-1-b)\ldots (-\omega^2-a)(-\omega^2-b)\ldots(-\omega -a)(-\omega -b)\ldots$$
$$=P(-1)P(-\omega^2)P(-\omega)$$
You’d then need to evaluate the product $P(-1)P(-\omega^2)P(-\omega) $. While it is a little tedious, it is doable in a reasonable amount of time, and it’s significantly faster than other methods.
A: The expression is the product of the roots of the quartic equation satisfied by $\,z = x^3 + 1\,$. Using polynomial resultants, the equation in $\,z\,$ is $\,\text{res}(x^4 - x^3 + 2 x^2 + x + 1, z - x^3 - 1, x)=0\,$ which results in $\,z^4 + 4 z^3 - 4 z^2 - 16 z + 16 = 0\,$, so the expression evaluates to $\,16\,$.
If not familiar with polynomial resultants, the same result can be derived as follows.

*

*Let $\,y = x^3\,$, then $\,x^4 = xy\,$ and $\,x^4 - x^3 + 2 x^2 + x + 1 = 0\,$ can be written as:
$$
2 x^2 + (y+1) x - z + 1= 0 \tag{1}
$$
Multiplying $\,(1)\,$ by $\,x\,$ and replacing $\,x^3 = y, \,x^4 = xy\,$ again:
$$
(y+1) x^2 - (y -1) x + 2 y = 0 \tag{2}
$$
Repeating one more time:
$$
(y-1) x^2 - 2 xy - y^2 - y = 0 \tag{3}
$$
Considering $\,(1), (2), (3)\,$ as a linear homogeneous system in $\,x^2, x, 1\,$, non-trivial solutions exist, so the determinant must be zero.

$$
0 = 
\begin{vmatrix}
   2 & y+1 & -y+1
\\ y+1 & -y+1 & 2 y
\\ y-1 & -2y & -y^2-y
\end{vmatrix}
= y^4 + 8 y^3 + 14 y^2 - 8 y + 1 \tag{4}
$$

*

*Let $\,z = y+1\,$, then substituting $\,y = z-1\,$ in $\,(4)\,$:
$$
\begin{align}
0 &= (z-1)^4 + 8 (z-1)^3 + 14 (z-1)^2 - 8 (z-1) + 1
\\ &= z^4 + 4 z^3 - 4 z^2 - 16 z + 16 \tag{5}
\end{align}
$$
A: 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.]
