A + B + C = 0, ABC finds a (B + C) + B (a + C) + C (a + b) ≠ abc≠0

A + B + C = 0, ABC finds a (B + C) + B (a + C) + C (a + b) ≠ abc≠0

a=b=1 c=-2
Then a (B + C) + B (a + C) + C (a + b) = - 6
a=1 b=2 c=-3
a(b+c)+b(a+c)+c(a+b)=-14
In fact, a (B + C) + B (a + C) + C (a + b) = - A ^ 2-B ^ 2-C ^ 2
So there are infinitely many conditional underscores

Elden Ring Premiere

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