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?

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.

#)Giải :

Ta có : \(\left(a+b+c\right)^3\)

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

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

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

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

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

Hay chính là \(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(\Rightarrowđpcm\)

ta có:

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

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

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

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

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

=>VT=VP( đpcm)

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

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

=a^3+b^3+c^3+3a^2c+6abc+3b^2c+3ac^2+3bc^2

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

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

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

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

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

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)