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分解因式:(1)x9+x6+x3-3;(2)(m2-1)(n2-1)+4mn;(3)(x+1)4+(x2-1)2+(x-1)4;(4)a3b-ab3+a2+b2+1.
题目详情
分解因式:
(1)x9+x6+x3-3;
(2)(m2-1)(n2-1)+4mn;
(3)(x+1)4+(x2-1)2+(x-1)4;
(4)a3b-ab3+a2+b2+1.
(1)x9+x6+x3-3;
(2)(m2-1)(n2-1)+4mn;
(3)(x+1)4+(x2-1)2+(x-1)4;
(4)a3b-ab3+a2+b2+1.
▼优质解答
答案和解析
(1)原式=x9+x6+x3-1-1-1
=(x9-1)+(x6-1)+(x3-1)
=(x3-1)(x6+x3+1)+(x3-1)(x3+1)+(x3-1)
=(x3-1)(x6+2x3+3)
=(x-1)(x2+x+1)(x6+2x3+3);
(2)原式=(m2-1)(n2-1)+2mn+2mn
=m2n2-m2-n2+1+2mn+2mn
=(m2n2+2mn+1)-(m2-2mn+n2)
=(mn+1)2-(m-n)2
=(mn+m-n+1)(mn-m+n+1);
(3)原式=(x+1)4+2(x2-1)2-(x2-1)2+(x-1)4
=[(x+1)4+2(x+1)2(x-1)2+(x-1)4]-(x2-1)2
=[(x+1)2+(x-1)2]2-(x2-1)2
=(2x2+2)2-(x2-1)2=(3x2+1)(x2+3);
(4)原式=a3b-ab3+a2+b2+1+ab-ab
=(a3b-ab3)+(a2-ab)+(ab+b2+1)
=ab(a+b)(a-b)+a(a-b)+(ab+b2+1)
=a(a-b)[b(a+b)+1]+(ab+b2+1)
=[a(a-b)+1](ab+b2+1)
=(a2-ab+1)(b2+ab+1).
=(x9-1)+(x6-1)+(x3-1)
=(x3-1)(x6+x3+1)+(x3-1)(x3+1)+(x3-1)
=(x3-1)(x6+2x3+3)
=(x-1)(x2+x+1)(x6+2x3+3);
(2)原式=(m2-1)(n2-1)+2mn+2mn
=m2n2-m2-n2+1+2mn+2mn
=(m2n2+2mn+1)-(m2-2mn+n2)
=(mn+1)2-(m-n)2
=(mn+m-n+1)(mn-m+n+1);
(3)原式=(x+1)4+2(x2-1)2-(x2-1)2+(x-1)4
=[(x+1)4+2(x+1)2(x-1)2+(x-1)4]-(x2-1)2
=[(x+1)2+(x-1)2]2-(x2-1)2
=(2x2+2)2-(x2-1)2=(3x2+1)(x2+3);
(4)原式=a3b-ab3+a2+b2+1+ab-ab
=(a3b-ab3)+(a2-ab)+(ab+b2+1)
=ab(a+b)(a-b)+a(a-b)+(ab+b2+1)
=a(a-b)[b(a+b)+1]+(ab+b2+1)
=[a(a-b)+1](ab+b2+1)
=(a2-ab+1)(b2+ab+1).
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