Given the terms of a arithmetic sequence a,b,c, verify if 3(a^2+b^2+c^2)=6(a-b)^2+(a+b+c)^2?

We notice that if we'll raise to square the sum a+b+c,
we'll get:

bc)

The identity to be verified
is:

= 6(a-b)^2+a^2 + b^2 + c^2
bc)

+ 2(ab
+ ac + bc)

+ 2(ab + ac +
bc)

We'll divide by 2 both
sides:

+ (ab + ac +
bc)

We'll expand the binomial from the right
side:

+ (ab + ac +
bc)

  = (ab + ac + bc) -
6ab

But b =

2b = a +
c

If we'll raise to square both sides, we'll
get:

c^2

6ab

6ab

6ab

6ab

-
12ab

c^2

d^2

4d^2

8ad + 4d^2 = 4a^2 + 4ad- a^2 - a^2- 4ad4d^2

8ad +
= -   -
8ad

We notice that the given expression does
not represent an identity since     , if a,b,c are the
consecutive terms of a arithmetical progression.