How do you simplify #(a-b)^2/(b-a)#?

1 Answer
May 11, 2016

Answer:

#(a-b)^2/(b-a) = b-a# excluding #a=b#

Explanation:

For any number #x# we have:

#(-x)^2 = x^2#

So:

#(a-b)^2/(b-a) = (-(a-b))^2/(b-a) = (b-a)^2/(b-a) = ((b-a)color(red)(cancel(color(black)((b-a)))))/color(red)(cancel(color(black)((b-a)))) = b-a#

excluding #a=b#

The exclusion is necessary because if #a=b# then #b-a = 0# resulting in a zero denominator in the expression:

#(a-b)^2/(b-a)#

and the result of division by #0# is undefined.