Pocket Aptitude

$x$ is clearly even.

$x\equiv 1^3+2^3+(-2)^3+(-1)^3\equiv 1^3+2^3-2^3-1^3\equiv 0\pmod{\! 5}$

$x\equiv 2^3+3^3+(-3)^3+(-2)^3\equiv 2^3+3^3-3^3-2^3\equiv 0\pmod{\! 7}$

$2,5,7\mid x\iff \text{lcm}(2,5,7)\mid x\iff 70\mid x\iff x\equiv 0\pmod{\! 70}$

by definition: $a\equiv b\pmod{\! m}\iff m\mid a-b$ .