2024/012) Given $a+b+c$ is divisible by 6 prove that $a^3+b^3+c^3$ is divisible by 6

We have $a^3-a= a(a^2-1) = a(a-1)(a+1) = (a-1)a(a+1)$ 

As $(a-1)a(a+1)$ is product of 3 consecutive  if is s divisible by 6 say 6m for some m

So $a^3 = a + 6m\cdots(1)$

Similarly 

$b^3 = b + 6p\cdots(2)$

and

$c^3 = c + 6q\cdots(3)$

Adding (1),(2) and (3) we get $a^3+b^3+c^3 = (a+b+c) +6(m+p+q)$

So if   $(a+b+c)$ is divisible by 6 then $a^3+b^3+c^3$ is divisible by 6