Đáp án+Giải thích các bước giải:
`C=a(b-c)^3+b(c-a)^3+c(a-b)^3`
`=a(b-c)^3-b(a-c)^3+c(a-b)^3`
Đặt `x=a-b;y=b-c⇒x+y=a-c` ta có:
`C=ay^3-b(x+y)^3+cx^3`
`=ay^3-b[x^3+y^3+3xy(x+y)]+cx^3`
`=ay^3-bx^3-by^3-3xy(x+y)b+cx^3`
`=(ay^3-by^3)-(bx^3-cx^3)-3xy(x+y)b`
`=y^3(a-b)-x^3(b-c)-3xy(x+y)b`
⇒ `C=(b-c)^3(a-b)-(a-b)^3(b-c)-3(a-b)(b-c)(a-c).b`
`=(a-b)(b-c)[(b-c)^2-(a-b)^2-3b(a-c)]`
`=(a-b)(b-c)(b^2-2bc+c^2-a^2+2ab-b^2-3ab+3bc)`
`=(a-b)(b-c)(c^2-a^2+bc-ab)`
`=(a-b)(b-c)[(c-a)(c+a)+b(c-a)]`
`=(a-b)(b-c)(c-a)(a+b+c)`
Vậy `C=(a-b)(b-c)(c-a)(a+b+c)`
`________________________`
`#Rùa~ ~ ~`
