C-03-Q101mediumsingle_mcqIf a+b+c=0a+b+c = 0a+b+c=0, a3+b3+c3=?a^3 + b^3 + c^3 = ?a3+b3+c3=?a3abc3abc3abcb000cabcabcabcd−3abc-3abc−3abcব্যাখ্যাWhen a+b+c=0a+b+c=0a+b+c=0, the identity a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca) has a zero right side, so a3+b3+c3−3abc=0a^3+b^3+c^3-3abc=0a3+b3+c3−3abc=0. Therefore a3+b3+c3=3abca^3+b^3+c^3=3abca3+b3+c3=3abc.