设(a2+b2)(c2+d2)=(ac+bd)2 ;+B,∵(a2+b2)(c2+d2)=a2c2+a2d2+b2c2+b2d2,(ac+bd)2 ;=a2c2+2abcd+b2d2,∴B=a2c2+a2d2+b2c2+b2d2-(a2c2+2abcd+b2d2),即B=a2d2-2abcd+b2c2=(ad-bc)2=(bc-ad)2.故答案是:ad-bc或bc-ad.