用简便方法计算:(a+b-c+d) * (a-b+c+d)

(a+b-c+d) * (a-b+c+d)
=[(a+d)+(b-c)][(a+d)-(b-c)]
=(a+d)²-(b-c)²
=a²+2ad+d²-(b²-2bc+c²)
=a²+2ad+d²-b²+2bc-c²

(a+b-c+d) * (a-b+c+d)
=[(a+d)+(b-c)][(a+d)-(b-c)]
=(a+d)^2-(b-c)^2
=a^2+2ad+d^2-b^2+2bc-c^2

令A=a+d,B=b-c原式可化为：
（A+B）(A-B) = A^2 - B^2,反代回去计算得：
a^2 + 2ad + d^2 - b^2 + 2bc - c^2

(a+b-c+d) * (a-b+c+d)=[(a+d)+(b-c)][(a+d)-(b-c)]
=(a+d)^2-(b-c)^2

原式=[(a+d)+(b-c)]*[(a+d)-(b-c)]
2 2
=(a+d) -(b-c)