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

Biến đổi VT ta có :

+) \(a^3+b^3+c^3=ab+bc+ca\)

\(\Leftrightarrow3a^3+3b^3+3c^3=3ab+3bc+3ca\)

\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=0\)

\(\Rightarrow a=b=c\)

< => VT = VP 

=> đpcm

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

                                                              \(=a^3+b^3=VT\)

ban su dung hang dang thuc la ra

Ta có : \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

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

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

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

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

Cách khác :

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

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^2-a^3-b^3-3a^2b-3ab^2=0\)

\(\Leftrightarrow0=0\left(luôn-đúng\right)\)

\(\Rightarrowđpcm\)

Ta có : \(VP=a^3-b^3-3ab\left(a-b\right)=a^3-b^3-3a^2b+3ab^2=\left(a-b\right)^3\)

=> \(\left(a-b\right)^3=a^3-b^3-3ab\left(a-b\right)\)

Vậy \(\left(a-b\right)^3=a^3-b^3-3ab\left(a-b\right)\).

Biến đổi vế trái:

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

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

Vậy VT = VP đẳng thức chứng minh

Biến đổi vế trái ta có:

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

\(VT=a^3-b^3-3ab.\left(a-b\right)=VP\) 

                                                      đpcm

a³-b³-3a²b-3ab²

=(a³-b³)=> đpcm

Sai đề chăng?

\(VT:\)\(\left(a-b\right)^3-a^3+b^3=a^3-3a^2b+3ab^2-b^3-a^3+b^3\)

\(=-3a^2b+3ab^2=-3ab\left(a-b\right)=VP\) (đpcm)

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

\(=a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2\)

\(=a^3+b^3=VT\)

p/s: chúc bạn học tốt

Ta có : \(\left(a+b\right)^3-3ab\left(a+b\right)=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2=a^3+b^3\)