PMO 3

Since $a$ , $b$ and $c$ are consecutive even numbers and $a>b>c$ , implies that $a = b+2$ and $c=b-2$ . Then, we have:

$\begin{aligned} x & = a^2+b^2+c^2 -ab-bc-ca \\ & = (b+2)^2 + b^2 + (b-2)^2 - (b+2)b-b(b-2) - (b-2)(b+2) \\ & = {\color{#3D99F6}b^2} + {\color{#3D99F6}2b} + 4 + {\color{#3D99F6}b^2} + {\color{#3D99F6}b^2} - {\color{#D61F06}2b} + 4 - {\color{#D61F06}b^2} - {\color{#D61F06}2b} - {\color{#D61F06}b^2} + {\color{#3D99F6}2b} - {\color{#D61F06}b^2} + 4 \\ & = \cancel{{\color{#3D99F6}b^2}} + \cancel{{\color{#3D99F6}2b}} + 4 + \cancel{{\color{#3D99F6}b^2}} + \cancel{{\color{#3D99F6}b^2}} - \cancel{{\color{#D61F06}2b}} + 4 - \cancel{{\color{#D61F06}b^2}} - \cancel{{\color{#D61F06}2b}} - \cancel{{\color{#D61F06}b^2}} + \cancel{{\color{#3D99F6}2b}} - \cancel{{\color{#D61F06}b^2}} + 4 \\ & = \boxed{12} \end{aligned}$

it's easy to calculate the result just by using 6 4 and 2 as a b and c, but the result is 12 for every triplets of consecutive even numbers. let's rewrite $a, b, c$ as $(n+4), (n+2), (n)$ .
the equation then becomes $(n+4)^2+(n+2)^2+n^2-n(n+4)-(n+4)(n+2)-n(n+2)$ which becomes $n^2+8n+16+n^2+4n+4+n^2-n^2-4n-n^2-6n-8-n^2-2n$ , all terms with $n$ or $n^2$ cancel out and we are left with $16+4-8=12$ .

b=a+2 c=a+4 therefore putting these values in equation we get 12