Factorising question help

It says to factorise this in my textbook:

(a + b)^2 − (b + c)^2

It says the answer is:

(a − c)(a + 2b + c)

Can someone please show full working and kind of explain how to get to the answer?

10 points for best answer.

The original expression is the difference of two squares

x² – y² = (x + y)(x – y)

(a + b + b + c)(a + b – b – c) (a + 2b + c)(a – c) as your book says.

Remember the difference of powers formula
a^2 - b^2 = (a - b)(a + b)

It's the same formula, just written differently.

(a + b)^2 - (b + c)^2 = ((a + b) - (b + c))((a + b) + (b + c)) = (a - c)(a + 2b + c)

You could multiply it out and re-factorise
or realise that this is a difference between two squares
So
(a + b)^2 − (b + c)^2
= (a + b + b + c)(a+b - b-c)
= (a-c)(a +2b +c)

(a + b)^2 − (b + c)^2 = a^2 + 2ab + b^2 - b^2 - 2bc - c^2
= (a^2 + 2ab) - (2bc - c^2)
because the b^2 terms cancel

The trick is to add in ac - ca = 0
(a^2 + 2ab + ac) - (ca -2bc - c^2)
= (a − c)(a + 2b + c)

Regards - Ian

Think of it as a difference of squares.

(a + b)^2 - (b + c)^2 = [(a + b) + (b + c)][(a + b) - (b + c)] = (a + 2b + c)(a - c)

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

(a + b)^2 − (b + c)^2
= [(a + b) + (b + c)][(a + b) - (b + c)]
= (a + 2b + c)(a - c)
= (a - c)(a + 2b + c)

(a + b)^2 − (b + c)^2 =
a^2+2ab+b^2 - b^2-c^2 -2bc =
a^2-c^2 +2b(a-c) =
(a+c)(a-c) + 2b(a-c) =
(a-c)(a+2b+c)