a4-b4+c4-d4-2(a2c2-b2d2)+4ac(b2+d2)-4bd(a2+c2)怎样分解因式

Question

a4-b4+c4-d4-2(a2c2-b2d2)+4ac(b2+d2)-4bd(a2+c2)怎样分解因式

Answer 1

a^4-b^4+c^4-d^4-2(a^2c^2-b^2d^2)+4ac(b^2+d^2)-4bd(a^2+c^2)
=a^4+2a^2c^2+c^4-4bd(a^2+c^2)+4b^2d^2-b^4-2b^2d^2-d^4+4ac(b^2+d^2)-4a^2c^2
=(a^2+c^2)^2-4bd(a^2+c^2)+(2bd)^2-[(b^2+d^2)^2-4ac(b^2+d^2)+(2ac)^2]
=[(a^2+c^2)-2bd]^2-[(b^2+d^2)-2ac]^2[平方差公式]
=[(a^2+c^2-2bd)+(b^2+d^2-2ac)][(a^2+c^2-2bd)-(b^2+d^2-2ac)]
=(a^2-2ac+c^2+b^2+d^2-2bd)(a^2+2ac+c^2-b^2-d^2-2bd)
=(a^2-2ac+c^2+b^2+d^2-2bd)[(a+c)^2-(b+d)^2]
=(a^2-2ac+c^2+b^2+d^2-2bd)(a+c+b+d)(a+c-b-d)