Ta có:
\(\left(a+b+c\right)^3\)
= \(\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)\)
= \(a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)\)
= \(a^3+b^3+3ab\left(a+b\right)+c^3+\left(3ac+3bc+3c^2\right)\left(a+b\right)\)
= \(a^3+b^3+c^3+\left(a+b\right)\left(3ab+3ac+3bc+3c^2\right)\)
= \(a^3+b^3+c^3+\left(a+b\right)[\left(3ab+3ac)+(3bc+3c^2\right)]\)
= \(a^3+b^3+c^3+\left(a+b\right)[3a\left(b+c)+3c(b+c\right)]\)
= \(a^3+b^3+c^3+\left(a+b\right)[\left(b+c\right)\left(3a+3c\right)]\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
(a+B+c)3=[(a+b)+C]3=(a+b)3+3(a+b)2c+3(a+b)c2+c3=a3+b3+3a2b+3ab2+3a2c+6abc+3b2c+3ac2+3bc2+c3
a3+b3+c3+3(a+b)(b+c)(c+a)=a3+b3+c3+6abc+3a2b+3ab2+3a2c+3b2c
+3ac2+3bc2.(nhân các đa thức 3(a+b)(a+c)(b+c) lại với nhau)
vậy (a+b+c)3=a3+b3+c3+3(a+b)(a+c)(b+c)
(a+b+c)3 = ((a+b)+c)3
= ( a+b)3+c3+3(a+b)2c+3(a+b)c2
= (a3+b3+3a2b+3ab2)+c3+3c(a2+b2+2ab)+3c2(a+b)
= a3+b3+c3+3(a2b+ab2+ca2+cb2+2abc+c2a+c2b)
= a3+b3+c3+3((ab+b2+ca+bc)a+(b2+ab+ca+cb)c)
= a3+b3+c3+3(ab+b2+ca+bc)(a+c)
= a3+b3+c3+3(a+b)+(b+c)+(a+c)