Prove Divisibility of $a^3+b^3+c^3$ Using $(a-b)^2+(b-c)^2+(c-a)^2=abc$

[anemone]

Let $a,\,b,\,c$ be integers such that $(a-b)^2+(b-c)^2+(c-a)^2=abc$. Prove that $a^3+b^3+c^3$ is divisible by $a+b+c+6$.

 

[kaliprasad]

Spoiler

we have $a^3+b^3+c^3-3abc$
=$\dfrac{1}{2}(a+b+c)(a^2+b^2+c^2 - ab - bc- ca)$
= $\dfrac{1}{2}(a+b+c)(2a^2+2b^2+2c^2 - 2ab - 2bc- 2ca)$
= $\dfrac{1}{2}(a+b+c)((a-b)^2 + (b-c)^2+ (c-a)^2)$
=$\dfrac{1}{2}(a+b+c)(abc)$

hence
$a^3+b^3+c^3 = \dfrac{1}{2}(a+b+c+6)(abc)$

now $(a+b+c+6)$ is a factor if $\dfrac{abc}{2}$ is integer

or atleast one of a,b,c is even.

all a,b,c cannot be odd then in the given condion
$(a-b)^2 + (b-c)^2 + (c-a)^2$ shall be even and abc shall be odd . so atleast one of a,b,c is even and so $\dfrac{abc}{2}$ is integer and hence given expression is divisible by $(a+b+c+6)$