Question

CMR: A³  + B³ + C³ -3ABC = 1/2(A+B+C)[(A-B)²  + (B-C)² + (A-C)² ]

Answer 1

Ta có : \(A^3+B^3+C^3-3ABC=\left(A+B\right)^3+C^3-3AB\left(A+B\right)-3ABC\)

\(=\left(A+B+C\right)\left[\left(A+B\right)^2-C\left(A+B\right)+C^2\right]-3AB\left(A+B+C\right)\)

\(=\left(A+B+C\right)\left(A^2+2AB+B^2-AC-BC+C^2-3AB\right)\)

\(=\left(A+B+C\right)\left(A^2+B^2+C^2-AB-BC-AC\right)\)

\(=\frac{A+B+C}{2}.\left[\left(A^2-2AB+B^2\right)+\left(B^2-2BC+C^2\right)+\left(C^2-2AC+A^2\right)\right]\)

\(=\frac{A+B+C}{2}\left[\left(A-B\right)^2+\left(B-C\right)^2+\left(C-A\right)^2\right]\)

Vậy ta có đpcm

 

Answer 2

Nhanh lên , gan đi học rồi