(v^2+1)^2-(v^2-2v-1)^2

Question

How can I solve this  (v^2+1)^2 - (v^2-2v-1)^2  and get 4v(v+1)(v-1) as a result? I'm solving the first part by applying a^2+2ab+b^2 and so far everything is fine, but then, I don't know what to do with the second part? Is there any equasion for squared trinomials that can be used, or the best way is to factorise the trinomial to a squared binomial? Please help and explai step by step. Thank you.

Answer 1

(v^2+1)^2-(v^2-2v-1)^2

Rule for squaring an expression with more than two terms.

Viz. Get the square of every term and two times all possible products.

e.g. (a+b+c+d)*2 = a^2 + b^2 + c^2 + d^2 + 2ab + 2ac  + 2ad + 2bc + 2bd + 2cd

So now square the two expressioms in the question. This gives us,

v^4 + 1 + 2v^2 - (v^4 + 4v^2 + 1 - 4v^3 - 2v^2 + 4v)

2v^2 - 4v^2 + 4v^3 + 2v^2 - 4v

4v^3 - 4v

4v(v^2 - 1)

4v(v + 1)(v - 1)