\displaystyle
\sum_{i\in\omega\cdot 3}(i^3+3^i) = \sum_{i\in\omega}(i^3+3^i) +
 \sum_{i\in\omega}\big((\omega+i)^3+3^{\omega+i}\big) +
 \sum_{i\in\omega}\big((\omega\cdot 2+ i)^3+3^{\omega\cdot 2+i}\big)