Is that not just (1 * (1-x^3) + x * (1-x^3) + x^2 * (1-x^3)) ??
The polynomial itself gives us the means at which we logically split up the multiplication into component parts.
Just as 3.14 * 3 == 3 * 3 + 0.1 * 3 + 0.04 * 3, when we move onto polynomials, we do the same exact thing. EDIT: remember, ALL REAL NUMBERS ARE POLYNOMIALS with a base of 10.
Or to put it another way: when x == 10, your polynomial of (1 + x + x^2) * (1-x^3) == 111 * (-999). That is to say: real numbers are simply polynomials where "x" has been defined to be a particular number, instead of an abstract entity. We call that number the radix-base.
If you instead defined the base to be x = 16 (hexadecimal numbers), you'd get 111 * (-FFF), which you'll find will satisfy similar properties. Now leave x-undefined (since it could be 10 or 16), and what do you get?
Polynomial math. Or so called "Carry-less multiplication" (https://en.wikipedia.org/wiki/Carry-less_product). We don't have a ring yet though: we still need to perform a modulus on all those polynomials to return to a proper ring (and if the modulus is irreducable, we have a Galois field). But we can already see how polynomials and the Reals are so closely related.