How do you factor
(a+b)^2-(2a-b)^2
and
(a+b)^2-4(b+c)^2
without simplifying the expression?
Each of them is a difference of two squares.
Remember that $\displaystyle (x + y)(x - y) = x^2 - y^2$.
$\displaystyle (a + b)^2 - (2a - b)^2 = [(a + b) + (2a - b)][(a + b) - (2a - b)]$
$\displaystyle = 3a(2b - a)$.
$\displaystyle (a+b)^2-4(b+c)^2 = (a + b)^2 - [2(b + c)]^2$
$\displaystyle = (a + b)^2 - (2b + 2c)^2$
$\displaystyle = [(a + b) + (2b + 2c)][(a + b) - (2b + 2c)]$
$\displaystyle = (a + 3b + 2c)(a - b - 2c)$.