Cube equal to a sum of three large cubes

Find the integer $x$ satisfying $x^3=441^3+588^3+735^3$.

$882$. The three numbers are $147\cdot3$, $147\cdot4$, $147\cdot5$, and $3^3+4^3+5^3=216=6^3$. So $x^3=147^3\cdot6^3$ and $x=882$.