Question

chứng minh hằng đẳng thức :

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

Answer 1

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