[Toán 8] PTĐTTNT

S

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Đặt $a-c=x \ \ \ \ c-b=y$

$\leftrightarrow (a-b)^5+(b-c)^5+(c-a)^5=(x+y)^5-
x^5-y^5=x^5+y^5+5x^4y+5xy^4+10x^2y^3+10x^3y^2-x^5-y^5$

$=5xy(x^3+y^3+2x^2y+2xy^2)=5xy[(x+y)^3-xy(x+y)]=5xy(x+y)
(x^2+xy+y^2)$

$=5(b-c)(c-a)(a-b)[(a-c)^2+(a-c)(c-b)+(c-b)^2]=5(b-c)(c-a)
(a-b)(a^2+b^2+c^2-ab-bc-ca)$

Vậy $(a-b)^5+(b-c)^5+(c-a)^5=5(a-b)(b-c)(c-a)(a^2+b^2+c^2-ab-bc-ca)$

 
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