Eighth grade fraction questions Given a + B-C / C = A-B + C / b = B + C-A / A, and ABC is not equal to 0, find the value of (a + b) (a + C) (B + C) / ABC X / Y Y is the denominator

Eighth grade fraction questions Given a + B-C / C = A-B + C / b = B + C-A / A, and ABC is not equal to 0, find the value of (a + b) (a + C) (B + C) / ABC X / Y Y is the denominator

From the known conditions: (a + b) / C-1 = (a + C) / B-1 = (B + a) / C-1
(a+b)/c=a+c/b=(b+a)c
Let K be their value, then CK = a + B, BK = a + C, CK = B + a
Three formula addition (a + B + C) k = 2 (a + B + C)
So a + B + C = 0 or K = 2
When a + B + C = 0, a + B = - C, a + C = - B, B + C = - A,
So the original formula = - ABC / ABC = - 1
When k = 2, a + B = 2c, a + C = 2B, B + C = 2A
So the original formula = 8abc / ABC = 8
To sum up, the original formula = - 1 or 8