有关因式分解的题求[(x^2+y^2)(a^2+b^2)+4abxy]^2-4[xy(a^2+b^2)+ab(x^2+y

[(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