[toán 8] khó

P

pinkylun

$a^3(c-b^2)+b^3(a-c^2)+c^3(b-a^2)+abc(abc-1)$

$=a^3c-a^3b^2+b^3a-b^3c^2+c^3b-c^3a^2+a^2b^2c^2-abc$

$=(a^2b^2c^2-c^3a^2-b^3c^2+bc^2)+(-a^3b^2+a^3c+b^3a-abc)$

$=c^2(a^2b^2-a^2c-b^3+bc)-a(a^2b^2-a^2c-b^3+bc)$

$=c^2(a^2-b)(b^2-c)-a(a^2-b)(b^2-c)$

$=(a^2-b)(b^2-c)(c^2-a)$

xong!1 :D :))
 
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