<s>[INST] Show that three numbers form an arithmetic progression 
The numbers $a,b$ and $c$ form  an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.

We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{   } \forall \text{   }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.
 [/INST] Expressions like $ \ a^2+ab+b^2 \  , \ a^2+ac+c^2 \ , \ b^2+bc+c^2 \ \ $  make me think of "differences of two cubes", so let's see what happens.  If we call the progression $ \ a \ = \ b - d \ , \ b \ , \ c \ = \ b + d \ \ , $ then we may write
$$ \ a^2 \ + \ ab \ + \ b^2 \ \ = \ \ \frac{b^3 \ - \ a^3}{b - a} \ \ = \ \ \frac{b^3 \ - \ (b - d)^3}{ d} \ \ = \  \  \frac{3·b^2d \ - \ 3·bd^2 \ + \ d^3}{d} \ \ , $$
$$ \ a^2 \ + \ ac \ + \ c^2 \ \ = \ \ \frac{c^3 \ - \ a^3}{c - a} \ \ = \ \ \frac{(b + d)^3 \ - \ (b - d)^3}{2d} \ \ = \  \  \frac{6·b^2d \  + \ 2·d^3}{2d} \ \ , $$
$$ \ b^2 \ + \ bc \ + \ c^2 \ \ = \ \ \frac{c^3 \ - \ b^3}{c - b} \ \ = \ \ \frac{(b + d)^3 \ - \ b^3}{d} \ \ = \  \  \frac{3·b^2d \ + \ 3·bd^2 \ + \ d^3}{d}  \ \ .  $$
The expressions in this sequence simplify to
$$ 3·b^2 \ - \ 3·bd \ + \ d^2 \ \ , \ \   3·b^2  \  + \ d^2 \ \ , \ \ 3·b^2  \ + \ 3·bd  \ + \ d^2 \ \ , $$
which differ sequentially by a constant amount $ \ 3·bd \ \ . $
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