Given a & sup2; + 2B & sup2; - 4A + 8b + 12 = 0, find a, B

Given a & sup2; + 2B & sup2; - 4A + 8b + 12 = 0, find a, B

a²+2b²-4a+8b+12=(a-2)²+2(b+2)²=0
Because (A-2) & sup2; ≥ 0, (B + 2) & sup2; ≥ 0
So only when (A-2) & sup2; = 0, (B + 2) & sup2; = 0 can the equation be satisfied
So a = 2, B = - 2

4a²b²-(a²+b²-c²)²

If the problem is factorization,
This is the 4A & sup2; a & sup2; B & sup2; (A & sup2; (A & sup2; + B & sup2; - C & sup2;) & sup2; (; (A & sup2; + B & sup2; - C & sup2; (; (A & sup2; + B & sup2; - C & sup2;) & sup2; (A & sup2; (A & sup2; + B & sup2; - C & sup2;) & sup2; = (2Ab) & sup2; (A & sup2; (A & sup2; (A & sup2; (A & sup2; (A & sup2; (A & B; (A & sup2; (a + b) & sup2; (A & sup2; (a + b) & amp; (A & sup2; (A & sup2; (A & sup2; (A & amp; (A & amp; (A & sup2; (A & amp; (A & amp; (A & sup2; (A & sup2; (A & amp; (A & sup2; (A & sup2; (A & amp; (a C) & sup2;

(A & sup2; + B & sup2;) & sup2; - 4A & sup2; B & sup2; simplification

(a²+1)²-4a²

(a²+1)²-4a²
=(a²+1-2a) (a²+1+2a)
=(a-1)²(a+1)²