Question

Chứng minh

x^3+y^3+z^3-3xyz=1/2(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]

Answer 1

x^3+y^3+z^3-3xyz=1/2(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]

2 cái bằng nhau

Answer 2

Chứng minh hộ tui phát

Answer 3

Ta có (a + b + c)³ =  a³ + b³ + c³ + 3a²b + 3a²c + 3b²a + 3b²c + 3c²a + 3c²b + 6abc

=> VT = (a + b + c)³ - (3a²b + 3a²c + 3b²a + 3b²c + 3c²a + 3c²b + 9abc)

= (a + b + c)³ - (3a²b + 3b²a + abc) - (3a²c + 3c²a + 3abc) - (3b²c + 3c²b + 3abc)

= (a + b + c)[a² + b² + c² + 2(ab + ac + bc) - 3(ab + bc + ac)]

= (a + b + c)(a² + b² + c² - ab - bc - ac)

VP = \(\frac{1}{2}\)(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]

= \(\frac{1}{2}\)(x+y+z)(2x² + 2b² + 2c² - 2ab - 2bc - 2ac)

= (x+y+z)(x² + b² + c² - ab - bc - ac)

Từ đó => VT=VP