Using modular polynomial arithmetic, perform the following multiplication over P

Irreducible polynomial is a polynomial that cannot be factorized into lower degree polynomials For the set of all polynomials over ${\displaystyle GF\left(2\right)}$, let's now consider polynomial arithmetic modulo the irreducible polynomial ${\displaystyle x^3+x+1}$.
When an algebric operation results in a polynomial whose degree equals or exceeds that of the irreducible polynomial, we will take for our result the remainder modulo the irreducible polynomial.
For example,
${\displaystyle (x^2+x+1)\times (x^2+1)b\text{mod}(x^3+x+1)}$
${\displaystyle =(x^4+x^3+x^2)+(x^2+x+1)b\text{mod}(x^3+x+1)}$
${\displaystyle =(x^4+x^3+x+1)b\text{mod}(x^3+x+1)}$
${\displaystyle =-x^2-x}$
${\displaystyle =x^2+x}$
Since ${\displaystyle 1+1=0}$ in GF(2). That's what caused the ${\displaystyle x^2}$ term to disappear in the second expression on the right hand side of the equality sign.
With multiplications modulo ${\displaystyle (x^3+x+1)}$, we have only the following eight polynomials in the set of polynomials over GF(2):
0
1
x
${\displaystyle x^2}$
${\displaystyle x+1}$
${\displaystyle x^2+1}$
${\displaystyle x^2+2}$
${\displaystyle x^2+x+1}$
We will refer to this set as ${\displaystyle GF\left(2^3\right)}$ where the exponent of 2, which in this case is 3, is the degree of the modulus polynomial.
We can think of xi as being merely a place-holder for a bit i.e., we can think of the polynomials as bit strings corresponding to the coefficients that can only be 0 or 1, each power of x representing a specific position in a bit string. So the ${\displaystyle 2^3}$ polynomials of ${\displaystyle GF\left(2^3\right)}$ can therefore be represented by the bit strings:
0----000
1-----001
x-----010
${\displaystyle x^2}$-----100
${\displaystyle x+1}$-----011
${\displaystyle x^2+1}$------101
${\displaystyle x^2+x}$----110
${\displaystyle x^2+x+1}$------111
If we wish, we can give a decimal representation to each of the above bit patterns. The decimal values between 0 and 7, both limits inclusive, would have to obey the addition and multiplication rules corresponding to the underlying finite field.