$\text{Đáp án:}$
Đặt a-b=x;b-c=y⇒x+y= a-b+b-c=a-c⇒c-a=-(x+y) khi đó thay vào ta được:
$$(a-b)^5+ (b-c)^5+ (c-a)^5$$
$$= x^5 + y^5 +[-(x+y)]^5$$
$$= x^5 + y^5 - (x+y)^5$$
$$= x^5 + y^5 - (x^5 + 5x^4y + 10x^3.y^2 + 10x^2.y^3 + 5xy^4 + y^5)$$
$$=x^5 + y^5 - x^5 - 5x^4y - 10x^3.y^2 - 10x^2.y^3 - 5xy^4 - y^5$$
$$= -( 5x^4y +10x^3.y^2 + 10x^2.y^3 + 5x.y^4)$$
$$=-5xy( x^3+ 2x^2.y + 2x.y^2 + y^3)$$
$$= - 5xy[(x+y)(x^2-xy+y^2) + 2xy(x+ y)]$$
$$= - 5xy(x+ y)(x^2+xy +y^2)$$
$$=-5(a-b)(b-c)[a^2-2ab+b^2+(a-b)(b-c)+b^2-2bc+c^2]$$
$$=-5(a-b)(b-c)(a^2+2b^2+c^2-2ab-2bc+ab-ac-b^2+bc)$$
$$=-5(a-b)(b-c)(a^2+b^2+c^2-ab-bc-ac)$$
