Đáp án:
$(a-c).(b-c).(a-b).(a+b+c)$
Giải thích các bước giải:
$ab.(a^2-b^2)+bc.(b^2-c^2)+ac.(c^2-a^2)$
$=b.[ a.(a^2-b^2)+c.(b^2-c^2)]+ac.(c^2-a^2)$
$=b.[a^3-ab^2+cb^2-c^3]+ac.(c^2-a^2)$
$=b.[ (a^3-c^3)-(ab^2-cb^2)]+ac.(c^2-a^2)$
$=b.[ (a-c).(a^2+ac+c^2)-b^2.(a-c)]+ac.(c^2-a^2)$
$=b. (a-c).(a^2+ac+c^2-b^2)-ac.(a^2-c^2)$
$=b.(a-c).(a^2+ac+c^2-b^2)-ac.(a-c).(a+c)$
$=(a-c).[ b.(a^2+ac+c^2-b^2)-ac.(a+c)]$
$=(a-c).[a^2b+abc+bc^2-b^3-a^2c-ac^2]$
$=(a-c).[(a^2b-a^2c)+(abc-ac^2)-(b^3-bc^2)]$
$=(a-c).[a^2.(b-c)+ac.(b-c)-b.(b-c).(b+c)]$
$=(a-c).(b-c).[(a^2+ac-b.(b+c)]$
$=(a-c).(b-c).[ a^2+ac-b^2-bc]$
$=(a-c).(b-c).[ (a^2-b^2)+(ac-bc)]$
$=(a-c).(b-c).[ (a-b).(a+b)+c.(a-b)]$
$=(a-c).(b-c).(a-b).(a+b+c)$
