Verify if (a²b-c)/(-c+ba²) = 1: Algebraic Fraction Analysis

Let's first examine the problem:

$\frac{a^2 b-c}{-c+b a^2}\stackrel{?}{= }1$Looking at the expression on the left side, note that using the distributive property in addition (first) and multiplication (later) in the fraction's denominator (or alternatively in the fraction's numerator) we get:

$\frac{a^2 b-c}{-c+b a^2}=\\ \frac{a^2 b-c}{b a^2+(-c)}=\\ \frac{a^2 b-c}{b a^2-c}=\\ \frac{a^2 b-c}{a^2b -c}=\\ 1$where in the final step we used the fact that dividing any number by itself always yields 1,

Therefore the expressions on both sides of the equality (which is assumed to hold) are indeed equal, meaning:

$\frac{a^2 b-c}{-c+b a^2}=\frac{a^2 b-c}{a^2b -c}\stackrel{!}{= }1$

(In other words, there is an identity equality- which is true for all possible values of the parameters $a,b,c$)

Therefore the correct answer is answer A.