Question

[Nguyễn Thị Kim Phương]

Chứng minh hằng đẳng thức:

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

Answer 1

[Rosan Yunaki]

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

= a³ + b³ + c³  + 3abc + 3ac² + 3b²c + 3bc² + 3a²b + 3a²c + 3b²a + 3abc

= ( a + b )³ + 3( a+b)²c + 3(a+b)c² + c³

= (a+b+c)³

Answer 2

[Trần Thanh Phương]

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

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

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

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

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

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

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

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

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