Question

因式分解 [(x^2+y^2)(a^2+b^2)+4abxy]^2-4[xy(a^2+b^2)+ab(x^2+y^2)]^2

Answer 1

[(x^2+y^2)(a^2+b^2)+4abxy]^2-4[xy(a^2+b^2)+ab(x^2+y^2)]^2
设x²+y²=m,a²+b²=n,xy=p,ab=q，则原式化为
（nm+4pq)²-4(pn+qm)²
=(nm+4pq+2pn+2qm)(nm+4pq-2pn-2qm)
=[n(m+2p)+2q(m+2p)][n(m-2p)-2q(m-2p)]
=(m+2p)(n+2q)(m-2p)(n-2q)
=(x²+y²+2xy)(a²+b²+2ab)(x²+y²-2xy)(a²+b²-2ab)
=(x+y)²（a+b)²（x-y)²（a-b)²
说明：换元法

Answer 2

实在是懒得打字，你看看吧。。。。。。其实我一直希望百度能给个公式编译器。。。。。。

Answer 3

[(x²+y²)(a²+b²)+4abxy]²-4[xy(a²+b²)+ab(x²+y²)]²
＝﹙a²x²＋b²x²＋a²y²＋b²y²＋4abxy﹚²－4﹙a²xy＋b²xy＋abx²＋aby²﹚²
＝[﹙a²x²＋2abxy＋b²y²﹚＋﹙b²x²＋2abxy＋a²y²﹚]²－4[﹙a²xy＋abx²﹚＋﹙b²xy＋aby²﹚]²
＝[﹙ax＋by﹚²＋﹙bx＋ay﹚²]²－4[ax﹙ay＋bx﹚＋by﹙bx＋ay﹚]²
＝[﹙ax＋by﹚²＋﹙bx＋ay﹚²]²－4[﹙ax＋by﹚﹙ay＋bx﹚]²
＝[﹙ax＋by﹚²＋﹙bx＋ay﹚²]²－[2﹙ax＋by﹚﹙ay＋bx﹚]²
＝[﹙ax＋by﹚²＋﹙bx＋ay﹚²＋2﹙ax＋by﹚﹙ay＋bx﹚][﹙ax＋by﹚²＋﹙bx＋ay﹚²－2﹙ax＋by﹚﹙ay＋bx﹚]
＝[﹙ax＋by﹚＋﹙bx＋ay﹚]²[﹙ax＋by﹚－﹙bx＋ay﹚]²
＝﹙ax＋by＋bx＋ay﹚²﹙ax＋by－bx－ay﹚²
＝[﹙ax＋bx﹚＋﹙ay＋by﹚]²[﹙ax－ay﹚－﹙bx－by﹚]²
＝[x﹙a＋b﹚＋y﹙a＋b﹚]²[a﹙x－y﹚－b﹙x－y﹚]²
＝[﹙a＋b﹚﹙x＋y﹚²[﹙x－y﹚﹙a－b﹚]²
＝﹙a＋b﹚²﹙x＋y﹚²﹙x－y﹚²﹙a－b﹚²

Answer 4

=(x²+y²+2xy)(a²+b²+2ab)(x²+y²-2xy)(a²+b²-2ab)
=(x+y)²（a+b)²（x-y)²（a-b)²