Answer:
Given:
[tex]\frac{(a-b)^2}{(b-a)^2}[/tex]
Solution:
Lets first expand the terms
[tex](a-b)^2 = a^2 +b^2 - 2ab[/tex]
[tex](b-a)^2 = a^2 +b^2 - 2ab[/tex]
Now
[tex]\frac{(a-b)^2}{(b-a)^2} = \frac{ a^2 +b^2 - 2ab}{ a^2 +b^2 - 2ab}[/tex]
Taking [tex]b^2[/tex]from both the numerator and denominator
[tex]\frac{(a-b)^2}{(b-a)^2} = \frac{ b^2(a^2 - 2ab)}{ b^2 (a^2 - 2ab)}[/tex]
[tex]\frac{(a-b)^2}{(b-a)^2} = \frac{(a^2 - 2ab)}{ (a^2 - 2ab)}[/tex]
Taking [tex]a^2[/tex] from both the numerator and denominator
[tex]\frac{(a-b)^2}{(b-a)^2} = \frac{ a^2( - 2ab)}{ a^2(- 2ab)}[/tex]
[tex]\frac{(a-b)^2}{(b-a)^2} = \frac{ ( - 2ab)}{ (- 2ab)}[/tex]
[tex]\frac{(a-b)^2}{(b-a)^2} = 1[/tex]
Thus [tex](a-b)^2 = (b-a)^2[/tex]
