Question 390900
You definitely mean {{{(a^2+b^2+c^2)/(a+b+c)=a-b+c}}}.

The given is {{{a/b = b/c}}}, that is, a,b,c are in continued proportion . One consequence of this condition is the fact that {{{ac = b^2}}}, which we will use later.

To start off, consider {{{(a+b+c)(a-b + c)}}}. Upon expansion, this is equal to {{{a^2 -ab + ac + ab - b^2 + bc + ac - bc + c^2 = a^2 + 2ac - b^2 + c^2}}}.
But  {{{ac = b^2}}}, so after substitution, 
{{{(a+b+c)(a -b + c) = a^2 + b^2 + c^2}}}.  Assuming that the sum a+b+c is not equal to zero, then we divide by a+b+c, and the result follows.

Consider a = 8, b = 4, and c = 2.  Then {{{(64+16 + 4)/(8+4+2) = 6}}}.