Question 3715
 |b+c a a|
 |b c+a b|
 |c c a+b|

 = (a+b)(b+c)(c+a) + 2abc - ca(c+a) - bc(b+c) - ab(a+b)
 = (b^2+ab+ca+bc)(c+a) + 2abc - ca(c+a) - bc(b+c) - ab(a+b)
 =  b^2c+abc+c^2a+bc^2 + ab^2+a^2b+ca^2+abc + 2abc - ca(c+a) - bc(b+c) - ab(a+b)
 =  (b^2c+ bc^2)+(c^2a+ ca^2) +(a^2b+ ab^2)+42abc - ca(c+a) - bc(b+c) - ab(a+b)
 = 4abc.

 Or let this determinant be f(a,b,c).

 We see that f(0,b,c) = (b+c) (cb - bc) = 0
 Similarly, f(a,0,c) = f(a,b,0) = 0
 
 Hence,abc is a divisor of f(a,b,c).
 Since deg f(a,b,c) = 3, assume f(a,b,c) = k abc.

 Set a=1,b=1,c =-1, we get
 |0  1  1|
 |1  0  1| = -1+(-1)-2 = -4 = k*1*1*(-1)= -k
 |-1 -1 2|

 So, k = 4 and f(a,b,c) = 4 abc.

 I prefer the 2nd method and hope that you can understand it.
 Kenny