Đáp án:
Bài `64: (b-c)^3+(c-a)^3+(a-b)^3`
`= b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3+a^3-3a^2b+3ab^2-b^3`
`= (b^3-b^3)-(c^3-c^3)-(a^3-a^3)-3(a^2b+ac^2-ab^2-bc^2+b^2c-a^2c)`
`=3[(a^2b-ab^2)+(ac^2-bc^2)-(a^2c-b^2c)]`
`=3[ab(a-b)+c^2(a-b)-c(a^2-b^2)`
`= 3[ab(a - b) + c^2(a - b) - c(a + b)(a - b)]`
`= 3(a - b)[ab + c^2 - c(a + b)]`
`= 3(a - b)(ab + c^2 - ac - bc)`
`= 3(a - b)(a - c)(b - c)`
