Toán Lớp 8: Chứng minh: a³+b³+c³=(a+b+c) (a²+b²+c² -ab-bc-ca)+3abc.

Question

Toán Lớp 8:  Chứng minh: a³+b³+c³=(a+b+c) (a²+b²+c²  -ab-bc-ca)+3abc.

bichhhathu · Answer

$  $    Giải đáp + giải th&iacute;ch c&aacute;c bước giải :    a^3 +b^3+c^3-3abc   = (a+b)^3 +c^3 - 3ab(a+b)-3abc   =(a+b+c)[(a+b)^2-(a+b)c+c^2] - 3ab(a+b+c)   = (a+b+c)(a^2+2ab+b^2 - ac-bc+c^2  -3ab)   =(a+b+c)(a^2 +b^2+c^2 - ab-bc-ca)   => a^3 +b^3+c^3-3abc = (a+b+c)(a^2 +b^2+c^2-ab-bc-ca)   => a^3 +b^3+c^3=(a+b+c)(a^2 +b^2+c^2-ab-bc-ca) +3abc (đpcm)

truongankimtrong · Answer

Giải đáp:Chứng minh: a&sup3;+b&sup3;+c&sup3;=(a+b+c) (a&sup2;+b&sup2;+c&sup2; -ab-bc-ca)+3abc.    => ta c&oacute;: a&sup3;+b&sup3;+c&sup3;=(a+b+c) (a&sup2;+b&sup2;+c&sup2; -ab-bc-ca)+3abc.    = (a+b+c).(a&sup2;+b&sup2;+c&sup2;-ab-bc-ac)+3.abc   =a.a&sup2;+a.b&sup2;+a.c&sup2;-a.ab-a.bc-a-ac+b.a&sup2;+b.b&sup2;+bc&sup2;-b.ab-b.bc-b.ac+c.a&sup2;+c.b&sup2;+c.c&sup2;-c.ab-c.bc-c.ac+3abc.   =a&sup3;+ab&sup2;+ac&sup2;-a&sup2;b-abc-a&sup2;c+a&sup2;b+b&sup3;+bc&sup2;-ab&sup2;-b&sup2;c-abc+a&sup2;c+b&sup2;c+c&sup3;-abc-bc&sup2;-ac&sup2;+3.abc.   =a&sup3;+(ab&sup2;-ab&sup2;) +(ac&sup2;-ac&sup2;)+(-a&sup2;b+ab)+(-abc-abc-abc)+(-a&sup2;c+a&sup2;c)+b&sup3;+(bc&sup2;-bc&sup2;)+(-b&sup2;c+b&sup2;c)+c&sup3;+3abc.   =a&sup3;-3abc+b&sup3;+c&sup3;+3abc.   = a&sup3;+(-3abc+3abc)+b&sup3;+c&sup3;.   =a&sup3;+b&sup3;+c&sup3;.   Vậy a&sup3;+b&sup3;+c&sup3;=(a+b+c).(a&sup2;+b&sup2;+c&sup2;-ab-bc-ac)+3bc.       Lời giải và giải thích chi tiết: