Let ABC not equal to 0, and find the value of fraction (a + b) (B + C) (c + a) / ABC. There are two answers, 8 and - 1. Let a + B-C / C = a + C-B / b = B + C-A / a = K

Let ABC not equal to 0, and find the value of fraction (a + b) (B + C) (c + a) / ABC. There are two answers, 8 and - 1. Let a + B-C / C = a + C-B / b = B + C-A / a = K

Add all the known numerator and denominator to get:
(a+b-c)/c=(a-b+c)/b=(-a+b+c)/a=(a+b+c)/(a+b+c)=1
a+b=2c,a+c=2b,b+c=2a,
So (a + b) (B + C) (c + a) / ABC
=2C * 2A * 2B / ABC = 8abc / ABC = 8 in conclusion, the values of (a + b) (B + C) (c + a) / ABC are - 1 and 8
Method 2: (a + b) / C-1 = (a + C) / B-1 = (B + C) / A-1
Subtract 1 at the same time, and you get
(a+b)/c=(a+c)/b=(b+c)/a=k
Let the above formula be equal to K, then a + B = KCa + C = kBb + C = Ka
2(a+b+c)=k(a+b+c)
k(a+b+c)-2(a+b+c)=0
(k-2)(a+b+c)=0
The solution is k = 2 and a + B + C = 0,
When k = 2, (a + b) (a + C) (B + C) / ABC = 2 * 2 * 2 = 8;
When a + B + C = 0, a + B = - C, a + C = - B, B + C = - A, then