3(1+a的平方+a的四次方)-(1+a+a的平方)的平方
=(3+3a^2+3a^4)-(a^4+2a^3+3a^2+2a+1)
=2a^4-2a^3-2a+2
=2(a^4-a^3-a+1)
=2(a-1)^2(a^2+a+1)
=2(a-1)^2[(a+1/2)^2+3/4]
3(1+a的平方+a的四次方)-(1+a+a的平方)的平方
=(3+3a^2+3a^4)-(a^4+2a^3+3a^2+2a+1)
=2a^4-2a^3-2a+2
=2(a^4-a^3-a+1)
=2(a-1)^2(a^2+a+1)
=2(a-1)^2[(a+1/2)^2+3/4]