If real numbers a, B and C satisfy a ^ 2 + B ^ 2 + C ^ x = 667, then the maximum value of algebraic formula (a-b) ^ 2 + (B-C) ^ 2 + (C-A) ^ 2 is

If real numbers a, B and C satisfy a ^ 2 + B ^ 2 + C ^ x = 667, then the maximum value of algebraic formula (a-b) ^ 2 + (B-C) ^ 2 + (C-A) ^ 2 is

(a-b)^2+(b-c)^2+(c-a)^2
=a^2+b^2+b^2+c^2+c^2+a^2-(2ab+2bc+2ac)
=2*667-(2ab+2bc+2ac)
And a ^ 2 + B ^ 2 > = 2Ab;
b^2+c^2>=2bc;
c^2+a^2>=2ac;
So 2Ab + 2BC + 2Ac

When a = - 1, the value of the algebraic formula (a + 1) 2 ← square + a (A-3) is equal to

A=-1
A+1=0
A-3=-1-3=-4
So (a + 1) ^ 2 + a (A-3)
=0^2+(-1)*(-4)
=0+4
=4

When a = - 1, the value of (a + 1) 2 + a (A-3) is ()
A. -4B. 4C. -2D. 2

According to the meaning of the question: (a + 1) 2 + a (A-3) = A2 + 2A + 1 + a2-3a = 2a2-a + 1, when a = - 1, the original formula = 2 × 1 + 1 + 1 = 4