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观察下列一组等式:
(a+1)(a2−a+1)=a3+1
(a−2)(a2+2a+4)=a3−8
(a+3)(a2−3a+9)=a3+27

(1)以上这些等式中,你有何发现?利用你的发现填空.
①(x-3)(x2+3x+9)=______;
②(2x+1)(______)=8x3+1;
③(______)(x2+xy+y2)=x3-y3
(2)计算:(a2-b2)(a2+ab+b2)(a2-ab+b2).
人气:257 ℃ 时间:2019-08-18 03:47:47
解答
(1)①(x-3)(x2+3x+9)=x3-27;
②(2x+1)(4x2-2x+1)=8x3+1;
③(x-y)(x2+xy+y2)=x3-y3
故答案为:①x3-27;②8x3+1;③x3-y3
(2)原式=[(a-b)(a2+ab+b2)][(a+b)(a2-ab+b2)]=(a3-b3)(a3+b3)=a6-b6
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