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 $\sum(ab)^3$?

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.
