If $a=2021$, $b=2022$, and $c=2023$, then the value of $a^3+b^3+c^3-3abc$ is:

We can use the well-known algebraic identity:
$$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$$

First, compute the differences:

$$a-b=2021-2022=-1$$

$$b-c=2022-2023=-1$$

$$c-a=2023-2021=2$$

Now, compute the sum of their squares:

$$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=(-1{)}^2+(-1{)}^2+(2{)}^2=1+1+4=6$$

Next, calculate the sum $a+b+c$:

$$a+b+c=2021+2022+2023=6066$$

Substitute these values back into the identity:

$$\text{Value}=\frac{1}{2}\times 6066\times 6=6066\times 3=18198$$