Factorisation, expanding brackets?

Question

Factorise:
(n² - a²) – (n – a)²

Answer 1

Some people have made mistakes in answering the question:
Here is how you factorise:
firstly, the first expression is the difference of two squares hence it can be written as: (n^2 - a^2) = (n - a)(n + a)
and the second expression can just be written as a normal breakdown of a square into its product of the same thing:
(n - a)^2 = (n - a)(n - a)
So, we now have the following:
(n^2 - a^2) - (n - a)^2
= (n - a)(n + a) - (n - a)(n - a)
So we can now take out the common factor, that being (n - a)
BUT be careful of the negative sign
= (n -a)[(n + a) - (n - a)]
= (n - a)[n + a - n + a]

So, the n's inside the square brackets cancel, giving
= (n - a)[2a]
= 2a(n - a)

and that's the answer

Answer 2

(n² - a²) – (n – a)²

(n² - a²) = (n + a)(n - a) {difference of two squares}

(n + a)(n - a) - (n - a)(n - a)
(n - a)[(n + a) - (n - a)]
(n - a)(n + a - n + a)
(n - a)(2a)
2a(n - a)

Answer 3

(n^2-a^2)-[(n-a)(n-a)]
(n^2-a^2)-[n^2-na-na+a^2]
(n^2-a^2)-[n^2-2na+a^2]
(n^2-a^2)+[-n^2+2na-a^2]
2na-2a^2

Answer 4

(n+a)(n-a)-(n-a)(n-a)
=(n-a)(n+a+n-a)
=2n(n-a)