Let $n=9867$. If you calculated $n^{3}-n^{2}$, what would be the unit digit found?
(a) 0
(b) 2
(c) 4
(d) 6
(e) 8
The correct option is (c).

Solution 1: The last digit of $9867^{3}$ is the same as that of $7^{3}=343$, which is 3. The last digit of $9867^{2}$ is the same as that of $7^{2}=49$, which is 9. If we subtract a number ending in 9 from another ending in 3, the last digit of the result is 4.

Note: Observe that the last digit of the difference $9867^{3}-9867^{2}$ is the same as the last digit of $\left(7^{3}-7^{2}\right)$.

Solution 2: $n^{3}-n^{2}=n^{2}(n-1)$, with $n^{2}=9867^{2}$ ending in 9 and $n-1=9866$ ending in 6. Since $9 \times 6=54$, the last digit of $n^{2}(n-1)$ is 4.