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
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