When to decide if cancellation is allowed?

[varun7nurav]

Hi guys,

Need help with a small doubt. I have seen conflicting solutions on Kaplan.
Problem: (x^2 - 1)/(x+1) = 1
In such situations, it looks evident to cancel out the common term after we expand the numerator which gives us the result (x-1)=1 resulting in x = 2.
Agree!
However, I have another problem that shows that such canceling isn't a good idea.
(a^2- b^2) = (a-b)^2
Solution 1: By expanding both sides, we get (a+b)(a-b)=(a-b)(a-b), canceling out common terms i.e. (a-b), we get (a+b)=(a-b), further solving gives us b=0 (Unique value)

Solution 2: By expanding using formula on RHS gives us
a^2 - b^2= (a^2 -2ab + b^2)
Canceling out a^2, we get b= 0 or b= a (Not unique)

This is something I really am not sure about.
Plz suggest me whr am I missing out? Am i forgetting any rule here?? If so, how do I decide when to cancel and when not to cancel out like terms?

Awaiting your response.
Thank you!

[pemdas]

you can't cancel out, unless you have sufficient data to set denominator different from 0

in your case Kaplan's quest. was quite deviant in canceling out denominator away
in the example (a^2- b^2) = (a-b)^2, you can cancel out only if (a-b)isn't equal to 0
then you are avoiding trivial a=b, because a-b=0 cannot be set as the required condition for canceling out. Your solution will be b=0, and a can be any value.

[varun7nurav]

Thank you!

[pavand]

You can cancel

In the second case, a^2-b^2 = (a+b)(a-b)
But (a-b)^2 ≠ (a-b)(a-b) because you are applying exponent before parentheses
(a+b) = (a-b) defies logic!

Think PEMDAS!

For example, (5-3)^2 = 2^2 (Parentheses) = 4 (Followed by Exponent)
Same as, 5^2 - 2*5*3 + 3^2 = 25 (E) - 30 (M) + 9 (E) = 4

5^2 - 3^2 = 25 - 9 = 16 ≠ (5-3)^2 = 4