Question

if a+b+c =abc,prove that (1+a^2)=(1-ab)(1-ac)​

Answer 1

Step-by-step explanation:

1/a^2 + 1/b^2 + 1/c^2 = 1/ab + 1/bc + 1/ca

1/a(1/a -1/b) + 1/b(1/b - 1/c) + 1/c(1/c -1/a) = 0

Multiplying by a^2b^2c^2

(b - a)/a^2b× a^2b^2c^2 = bc^2(b -a)

(c - b)/b^2c × a^2b^2 c2 = a^2c(c - b)

(a - c )/c^2a × a^2b^2c^2 = ab^2( a - c)

bc^2(b- a) +a^2c (c - b) + ab^2(a - c) = 0

This equation entails that this is true only when

b - a = 0. Or. b = a

c - b = 0. Or c = b

a - c = 0. Or. a = c

Therefore

a = b= c