i think this question goes well with good wine

how should i know what value is b?

start plugging -1 ---> i get 1 =-b+b so how can understand if \(a≠b\) cant understand the concept ..

=>

\(\frac{a^2}{(a-b)}=b\)
\(⇔ a^2=b(a-b)\)
\(⇔ a^2=ab – b^2\)
\(⇔ a^2 - ab + b^2 = 0\)
Since \(a≠b\), one of a and b is not zero.
If \(a ≠ 0\) or \(b ≠ 0\), \(a^2 - ab + b^2\) cannot be zero for the following reason:
\(a^2 - ab + b^2 = a^2 – 2a(\frac{b}{2}) + b^2 = a2 – 2a(\frac{b}{2}) + \frac{b^2}{3} + (\frac{3}{4})b^2\)
\(= (a – \frac{b}{2})^2 + (\frac{3}{4})b^2 > 0\)
Thus, there is no pair of real numbers \((a,b)\) satisfying the equation \(a^2 - ab + b^2 = 0.\)

Therefore, \(a\) cannot exist, and the answer is E.
Answer: E
_________________