Question

(a+b+c)×(b+c-a)×(c+a-b)×(a+b-c)

Answer 1

a + b + c = 1

Thus, a + b = 1 - c

b + c = 1 - a

c + a = 1 - b

(1/a) + (1/b) + (1/c) = 3

Multiplying both sides by (abc), we get,

1bc/abc + 1ca/bca + 1ab/cab = 3

(ab + bc + ca)/abc = 3

ab + bc + ca = 3abc

Now,

(a + b)ab + (b + c)bc + (c + a)ca

= (1 - c)ab + (1 - a)bc + (1 - b)ca

= ab + bc + ca - 3(abc)

= 3abc - 3abc

= 0

Therefore, (a + b)ab + (b + c)bc + (c + a)ca = 0