Comment: Note that:

the units digit of $\left(9867^{3}-9867^{2}\right)=$ the units digit of $\left(7^{3}-7^{2}\right)$
2: $n^{3}-n^{2}=n^{2}(n-1)$. Thus, $n^{2}=(9867)^{2}$ ends in 9 and $n-1=9866$ ends in 6. Since, $9 \times 6=54$, the last digit of the result is 4.