5. If $m=1996^{3}-1995^{3}+1994^{3}-1993^{3}$ $+\cdots+4^{3}-3^{3}+2^{3}-1^{3}$, then the last digit of $m$ is
5. (0).
$$
2^{3}-1^{3}=7,4^{3}-3^{3}=37,6^{3}-5^{3}=91,8^{3}-7^{3}=
$$
$169,10^{3}-9^{3}=271$. Therefore, the sum of the last digits of the first 10 numbers $10^{3}-9^{3}+8^{3}-7^{3}$ $+6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}$ is $1+9+1+$ $7+7=25$, and the last digit of 25 is 5. The last digit of the algebraic sum of every ten consecutive cubic numbers from the unit digit 1 to the unit digit 0 is always 5. For $1990^{3}$, there are a total of 199 fives, and the last digit remains 5. Additionally, $1996^{3}-1995^{3}+1994^{3}-1093^{3}+1992^{3}-1591^{3}$ also has a last digit of 5. Therefore, the last digit of $m$ is ).