Prove that the terms x+1, x^2-1, x^3-x^2-x+1 are the terms of a G.P.

To prove that the given terms are the consecutive terms of
a geometric series, we'll use the constraint imposed to the middle term of 3 consecutive
terms of a geometric series.

The middle terms is  x^2-1 and
it has to be the geometric mean of the neighbor
terms.

 x^2-1 = sqrt[(x+1)*(x^3-x^2-x+1)]
(1)

We notice that if we factorize last term x^3-x^2-x+1,
we'll get:

x^2(x-1) -
(x-1)

We'll factorize
again:

x^2(x-1) - (x-1) = (x-1)(x^2 - 1)
(2)

The factor x^2 - 1 is a difference of
squares:

x^2 - 1 = (x-1)(x+1)
(3)

We'll substitute (3) in
(2):

x^2(x-1) - (x-1) =
(x-1)(x-1)(x+1)

x^2(x-1) - (x-1) = (x-1)^2*(x+1)
(4)

We'll substitute (4) in
(1):

 x^2-1 =
sqrt[(x+1)*(x-1)^2*(x+1)]

 x^2-1 =
sqrt[(x+1)^2*(x-1)^2]

 x^2-1 =
(x+1)(x-1)

We have obtained an identity, so the given terms
are the consecutive terms of a geometric progression.