Given: (b + c – a)x = (c + a – b)y = (a + b – c)z = 2

x = 2\over \text{b + c - a} or 1/x = \text{b + c - a}\over 2

y = 2\over \text{c + a - b} or, 1/y = \text{c + a - b}\over 2

z = 2\over \text{a + b - c} or, 1/z = \text{a + b - c}\over 2

Calculating (1/x + 1/y):

(1/x + 1/y) = \text{b + c - a}\over 2 + \text{c + a - b}\over 2

= \text{b + c - a + c + a - b}\over 2

= c

Calculating (1/y + 1/z):

(1/y + 1/z) = \text{c + a - b}\over 2 + \text{a + b - c}\over 2

= \text{c + a - b + a + b - c}\over 2

= a

Calculating (1/z + 1/x):

(1/z + 1/x) = \text{a + b - c}\over 2  + \text{b + c - a}\over 2

= \text{a + b - c + b + c - a}\over 2

= b

LHS: (1/x + 1/y)(1/y + 1/z)(1/z + 1/x)

= c × b × a

= abc RHS