Question 944201
<pre>
Since a,b,c are in H.P., then 1/a,1/b,1/c are in A.P., so 1/b-1/a=1/c-1/b,
or 1/c = 2/b-1/a, substituting that for 1/c in

 (1/a + 1/b - 1/c)(1/b + 1/c - 1/a) 

ans simplifying, gives

((2b-a)(3a-2b))/(a^2b^2)

which is nothing special at all.

Edwin</pre>