Ta có \(VT=\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+3\left(a+b\right)^2.c+3\left(a+b\right)c^2+c^3\)

\(=a^3+3a^2b+3ab^2+b^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left[\left(a+b\right)c+c^2+ab\right]\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)\right]+c\left(b+c\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)

Vậy \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

(a+b+c)^3=((a+b)+c)^3=(a+b)^3+c^3+3(a+b)c(a+b+c)
=a^3+b^3+3ab(a+b)+c^3+3(a+b)c(a+b+c)
=a^3+b^3+c^3+3(a+b)(ab+c(a+b+c))
=a^3+b^3+c^3+3(a+b)(ab+ac+bc+c^2)
=a^3+b^3+c^3+3(a+b)(a+c)(b+c)

có tính chất (a+b)ⁿ=aⁿ+bⁿ à.nếu có chứng minh?

(a+b+c)³=[(a+b)+c]³=(a+b)³+c³+3(a+b)c(a+b+c)
=a³+b³+3ab(a+b)+c³+3(a+b)c(a+b+c)
=a³+b³+c³+3(a+b)[ab+c(a+b+c)]
=a³+b³+c³+3(a+b)(ab+ac+bc+c2)

==a³+b³+c³+3(a+b)[(ab+ac)+(bc+c2)]

=a³+b³+c³+3(a+b)(a+c)(b+c)

#)Giải :

\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+ca+c^2\right)\)

\(=a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)\)

\(=\left(a+b^3\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)\)

\(=\left(a+b+c\right)^3\)

\(\Rightarrowđpcm\)

Biến đổi vế trái:

a + b + c 3 = a + + c 3  = a + b 3 +3 a + b 2  c+3(a+b) c 2 + c 3

      =  a 3  + 3 a 2 b + 3a b 2  +  b 3  + 3( a 2 + 2ab +  b 2 )c + 3a c 2  + 3b c 2  +  c 3

      =  a 3  + 3 a 2 b + 3a b 2  +  b 3  + 3 a 2 c + 6abc + 3 b 2 c + 3a c 2  + 3b c 2 + c3

      =  a 3 +  b 3  +  c 3  + 3 a 2 b + 3a b 2 + 3 a 2 c + 6abc + 3 b 2 c + 3a c 2  + 3b c 2

      =  a 3  +  b 3  +  c 3  + (3 a 2 b + 3a b 2 ) +( 3 a 2 c + 3abc)+ (3abc + 3 b 2 c)+(3a c 2  + 3b c 2 )

      =  a 3  +  b 3  +  c 3  + 3ab(a + b) + 3ac(a + b) + 3bc(a + b) + 3 c 2 (a + b)

      =  a 3  +  b 3  +  c 3 + 3(a + b)(ab + ac + bc +  c 2 )

      =  a 3  +  b 3  +  c 3  + 3(a + b)[a(b + c) + c(b + c)]

      =  a 3  +  b 3  +  c 3  + 3(a + b)(b + c)(a + c) (đpcm)

Để chứng minh hằng đẳng thức a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a) = (a+b+c)^3, ta sẽ sử dụng công thức khai triển đa thức.

Theo công thức khai triển đa thức, ta có:

(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)

Vậy, hằng đẳng thức được chứng minh.

(a+b+c)^3=((a+b)+c)^3=(a+b)^3+c^3+3(a+b)c(a+b+c)
=a^3+b^3+3ab(a+b)+c^3+3(a+b)c(a+b+c)
=a^3+b^3+c^3+3(a+b)(ab+c(a+b+c))
=a^3+b^3+c^3+3(a+b)(ab+ac+bc+c^2)
=a^3+b^3+c^3+3(a+b)(a+c)(b+c)

(a+b+c)³=((a+b)+c)³=(a+b)³+c³+3(a+b)c(a+b+c)

=a³+b³+3ab(a+b)+c³+3(a+b)c(a+b+c)

=a³+b³+c³+3(a+b)(ab+c(a+b+c))

=a³+b³+c³+3(a+b)(ab+ac+bc+c²)

=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)c(a+b+c)

=a³+b³+3ab(a+b)+c³+3(a+b)c(a+b+c)

=a³+b³+c³+3(a+b)(ab+c(a+b+c))

=a³+b³+c³+3(a+b)(ab+ac+bc+c²)

=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)c(a+b+c)

=a³+b³+3ab(a+b)+c³+3(a+b)c(a+b+c)

=a³+b³+c³+3(a+b)(ab+c(a+b+c))

=a³+b³+c³+3(a+b)(ab+ac+bc+c²)

=a³+b³+c³+3(a+b)(a+c)(b+c)

a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

=> ĐPCM

b) \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)^2+c^3-3abc\)

\(=\left[\left(a+b\right)^3+c^3\right]-\left(3a^2b+3abc+3ab^2\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2-3ab\right]\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

=> ĐPCM

P/s: Có sao sót xin bỏ qua

a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+3\left(a+b\right)^2\cdot c+3\left(a+b\right)c^2+c^3\)\(-a^3-b^3-c^3\)

\(=a^3+b^3+c^3+3a^2b+3ab^2+3\left(a^2+2ab+b^2\right)c\)\(+3ac^2+3bc^2-a^3-b^3-c^3\)

\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)

\(=\left(3abc+3a^2c+3b^2c+3bc^2\right)\)\(+\left(3a^2b+3a^2c+3ab^2+3abc\right)\)

\(=c\left(3ab+3ac+3b^2+3bc\right)\)\(+a\left(3ab+3ac+3b^2+3bc\right)\)

\(=\left(a+c\right)\left[\left(3ab+3b^2\right)+\left(3ac+3bc\right)\right]\)

\(=\left(a+c\right)\left[3b\left(a+b\right)+3c\left(a+b\right)\right]\)

\(=3\left(a+c\right)\left(a+b\right)\left(b+c\right)\)

b) \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)( do \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\))

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)\(-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ab-ac\right)\)\(-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)