= [(a+b+c)³ - a³] - [b³ + c³]
= [(a+b+c-a)((a+b+c)² + a(a+b+c) + a²)] - [(b+c)(b² - bc + c²)]
= [(b+c)(a² + b² + c² + 2ab + 2bc + 2ca + a² + ab + ac + a²)] - [(b+c)(b² - bc + c²)]
= (b+c)[3a² + 3ab + 3ac + 3bc]
= 3(b+c)[a(a+b) + c(a+b)]
= 3(b+c)(a+b)(a+c)
= 3(a+b)(b+c)(c+a)
अतः, (a+b+c)³ - a³ - b³ - c³ = 3(a+b)(b+c)(c+a) सिद्ध हुआ।
