一道初一数学题!!!急

a^4+2a^3+2a-1
=a^4-1+2a^3+2a
=(a^2+1)(a^2-1)+2a(a^2+1)
=(a^2+1)(a^2-1+2a)
因为a^2+2a=1
所以a^2-1+2a=0
所以(a^2+1)(a^2-1+2a)=0
所以a^4+2a^3+2a-1=0

a的2次方＋2a＝1

a的4次方＋2a的3次方＋2a－1
=a^2（a^2+2a）+2a-1
=a^2+2a-1
=1-1
=0

a的2次方＋2a＝1
即a^2+2a-1=0

对a的4次方＋2a的3次方＋2a－1进行因式分解并求值：
a^4+2a^3+2a-1
=a^4-1+2a(a^2+1)
=(a^2-1)(a^2+1)+2a(a^2+1)
=(a^2+1)(a^2-1+2a)
=0

a^4+2a^3+2a-1
=a^2(a^2+2a)+2a-1
=a^2+2a-1
=1-1
=0