Question

Cm đẳng thức:(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)

Answer 1

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

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

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

\(=\left(a^3+b^3+c^3+3abc+3a^2b+3ac^2+3a^2c+3ab^2+3bc^2\cdot3bc^2+3abc\right)-\\ \left(a^3+3a^2b+3ab^2+b^3+3a^2c+6abc+3b^2c+3ac^2+3bc^2+c^3\right)\)

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

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

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

\(=0\)

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

Answer 2

https://olm.vn/hoi-dap/question/161510.html