Knowing that I don't see an equal sign on this polynomial, it is meant that it must be treated as an expression, so I assume the objective with this expression is to factor it.
If factoring is the objective in this exercise, consider there are two areas of interest in this expression:
=> First, c2 - 2c + 1 is a perfect square trinomial, so it is factored this way:
c2 - 2c + 1 = c2 - c - c + 1
= c(c - 1) - 1(c - 1)
= (c - 1)(c - 1)
= (c - 1)2
Note: The above trinomial was factored by grouping.
Second, the term 9f 2 can be written as (3f)2 knowing that both 9 and f 2 are squares.
With both areas of interest being covered, the original expression can be rewritten as (c - 1)2 - (3f )2.
Notice that the rewritten expression has the following notation: x2 - y2, which is a difference of squares. In this case, x = (c - 1)2 and y = (3f)2.
Since, for any difference of squares, x2 - y2 = (x - y)(x + y), the expression
(c - 1)2 - (3f)2 = [(c - 1) - 3f] [(c - 1) + 3f].
The use of brackets was needed to distinguish one factor from the other.