Question

chứng minh 2(a³+b³+c³-3abc)=(a+b+c)[(a-b)²+(b-c)²+(c-a)² ]

Answer 1

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

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

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

Answer 2

Ta có : 2 ( a^3 + b^3 + c^3 - 3abc ) = 2 ( a + b + c ) ( a^2 + b^2 + c^2 - ab - ac - bc )

= ( a + b + c ) ( 2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2 bc )

= ( a + b + c ) [ ( a - b )^2 + ( b-c )^2 + ( c - a )^2 ] ( đpcm )