(a+b+c)3−a3−b3−c3(a+b+c)3−a3−b3−c3
=a3+3a2(b+c)+3a(b+c)2+(b+c)3−a3−b3−c3=a3+3a2(b+c)+3a(b+c)2+(b+c)3−a3−b3−c3
=3(b+c)(a2+ab+ac)+b3+3b2c+3bc2+c3−b3−c3=3(b+c)(a2+ab+ac)+b3+3b2c+3bc2+c3−b3−c3
=3(b+c)(a2+ab+ac+bc)=3(b+c)(a2+ab+ac+bc)
=3(b+c)[a(a+b)+c(a+b)]=3(b+c)[a(a+b)+c(a+b)]
=3(b+c)(a+b)(a+c)