jmc

To find the sum, we compute the first few cubes modulo 6: $$\begin{array}{rl}{1}^3& \equiv 1,\\ 2^3& \equiv 8\equiv 2,\\ 3^3& \equiv 27\equiv 3,\\ 4^3& \equiv 64\equiv 4,\\ 5^3& \equiv 125\equiv 5,\\ 6^3& \equiv 0(\mathrm{mod}6).\end{array}$$We see that $n^3\equiv n(\mathrm{mod}6)$ for all integers $n$, so $$\begin{array}{rl}{1}^3+2^3+3^3+\cdots +100^3& \equiv 1+2+3+\cdots +100\\ & \equiv \frac{100\cdot 101}{2}\\ & \equiv 5050\\ & \equiv \boxed{4}(\mathrm{mod}6).\end{array}$$