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 $\sum(ab)^3$?

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 $\,(4)\,$:
$$
\begin{align}
0 &= (z-1)^4 + 8 (z-1)^3 + 14 (z-1)^2 - 8 (z-1) + 1
\\ &= z^4 + 4 z^3 - 4 z^2 - 16 z + 16 \tag{5}
\end{align}
$$
