Problem: Reduced to lowest terms. a2−b2ab−ab−b2ab−a2\dfrac{a^{2}-b^{2}}{a b}-\dfrac{a b-b^{2}}{a b-a^{2}}aba2−b2−ab−a2ab−b2 is equal to:
Answer Choices:
A. ab\dfrac{a}{b}ba
B. a2−2b2ab\dfrac{a^{2}-2 b^{2}}{a b}aba2−2b2
C. a2a^{2}a2
D. a−2ba-2 ba−2b
E. none of these answers. Solution:
a2−b2ab−b(a−b)a(b−a)=a2−b2ab+b2ab=ab\dfrac{a^{2}-b^{2}}{a b}-\dfrac{b(a-b)}{a(b-a)}=\dfrac{a^{2}-b^{2}}{a b}+\dfrac{b^{2}}{a b}=\dfrac{a}{b}aba2−b2−a(b−a)b(a−b)=aba2−b2+abb2=ba.