A is not B

$\frac{a^{2}-b^{2}}{a-b}$ can be expressed as $\frac{(a-b)(a+b)}{a-b}$ by factoring the top and bottom cancel out leaving $a+b$

can be expressed as by factoring the top and bottom cancel out leaving

very very very easy

$\frac{a^2-b^2}{a-b}=\frac{(a-b)(a+b)}{a-b}=\boxed{\large{a+b}}$

This sum is related to identity (a+b)(a-b).After that we can write (a+b)(a-b)/(a-b). And by factorizing it, the answer will be (a+b).

(a^2-b^2)/(a-b) can be written as (a+b)(a-b)/(a-b). So the answer is a+b

Just use a^2-b^2= a+b a-b

Just require an inspection calculation.

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

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

difference of two perfect squares then cancel like terms....

{ a }^{ 2 } - { b }^{ 2 } = (a-b) * (a+b)

so numerator a-b divide by a-b: a+b

Easy as any expansion gets.