समीकरण (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0
A = c2 – ab, B = -2(a2 – bc), C = b2 – ac
∵ समीकरण के मूल बराबर हैं।
∴ (B)2 – 4AC = 0
[- 2(a2 – bc)]2 – 4(c2 – ab)(b2 – ac) = 0
4(a2 – bc)2 – 4(b2c2 – ac3 – ab3 + a2bc) = 0
4(a4 + b2c2 – 2a2bc – b2c2 + ac3 + ab3 – a2bc) = 0
a4 + ab3 + ac3 – 3a2bc = 0
a(a3 + b3 + c3 – 3abc) = 0
a = 0 तथा a3 + b3 + c3 – 3abc = 0
a3 + b3 + c3 = 3abc
अतः a = 0 या a3 + b3 + c3 = 3abc इति सिद्धम्।
