If $a=999$, $b=998$, $c=997$, then the value of $a^3+b^3+c^3-3abc$ will be:

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

Given $a=999$, $b=998$, $c=997$:

$$a+b+c=999+998+997=2994$$

$$a-b=999-998=1$$

$$b-c=998-997=1$$

$$c-a=997-999=-2$$

Substitute these values into the formula:

$$\text{Value}=\frac{1}{2}(2994)\left[1^2+1^2+(-2{)}^2\right]$$

$$\text{Value}=\frac{1}{2}(2994)[1+1+4]=\frac{1}{2}(2994)[6]=2994\times 3=8982$$