Remainder of sum of cubes divided by 70

$x = 16^3 + 17^3 + 18^3 + 19^3$ is an even number, so $2$ divides $x$. Since $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, $a + b$ always divides $a^3 + b^3$. Therefore, $16^3 + 19^3$ is divisible by $16 + 19 = 35$, and $17^3 + 18^3$ is divisible by $17 + 18 = 35$. Thus, $x$ is divisible by $35$. Since $x$ is also even, $x$ is divisible by $70$. Hence, the remainder when $x$ is divided by $70$ is $0$.