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:

`(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + ac +
bc)`

The identity to be verified
is:

`3(a^2+b^2+c^2)=6(a-b)^2+(a+b+c)^2`

`3(a^2+b^2+c^2)
= 6(a-b)^2+a^2 + b^2 + c^2`  + 2(ab + ac +
bc)

`3(a^2+b^2+c^2) - (a^2 + b^2 + c^2) = 6(a-b)^2` + 2(ab
+ ac + bc)

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

We'll divide by 2 both
sides:

`(a^2+b^2+c^2) = 3(a-b)^2` + (ab + ac +
bc)

We'll expand the binomial from the right
side:

`(a^2+b^2+c^2) = 3a^2 - 6ab + 3b^2` + (ab + ac +
bc)

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

But b = `(a+c)/2`

2b = a +
c

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

`4b^2 = a^2 + 2ac + c^2 =gt 2ac = 4b^2 - a^2 -
c^2`

`-2a^2 - 2b^2+ c^2 = b(a+c) + ac -
6ab`

`-2a^2 - 2b^2+ c^2 = 2b^2 + (4b^2 - a^2 - c^2)/2 -
6ab`

`-2a^2 - 2b^2+ c^2 = 4b^2 - a^2/2 - c^2/2 -
6ab`

`-6b^2 = 3a^2/2 + 3c^2/2 -
6ab`

-`12b^2 = 3(a^2 + c^2) -
12ab`

`4b^2 = 4ab - a^2 -
c^2`

`b = a+d =gt b^2 = a^2 + 2ad +
d^2`

`c = a+2d =gt c^2 = a^2 + 4ad +
4d^2`

`ab = a^2 + ad`

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

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

We notice that the given expression does
not represent an identity since `8d^2`  `!=`  `-2a^2 - 8ad` , if a,b,c are the
consecutive terms of a arithmetical progression.