2014/112) if x+y is divisble by 3 show that $x^3+y^3$ is divisible by 9

we have
$x^3+y^3= (x+y)^3 - 3xy(x+y)$
if $(x+y)$ is divisible by 3 then $(x+y)^3$ is divisible by 9 ( it is divisible by 27 but 9 is required) and $3xy(x+y)$ is divisible by 9 and hence the difference.