Solution
= Solution
Use <modular arithmetic> to keep all intermediate numbers small:
$$
34\equiv2,\quad45\equiv5,\quad55\equiv7,\quad89\equiv1\pmod8.
$$
Thus $34\cdot45\cdot55+89\equiv2\cdot5\cdot7+1=71\equiv7\equiv-1\pmod8$. Cubing gives
$$
\boxed{(34\cdot45\cdot55+89)^3\equiv(-1)^3\equiv7\pmod8;\quad\text{remainder }7.}
$$