Question

因式分解：x的8次方+x的6次方+x的4次方+x的2次方+1

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Answer 1

(x^4+x^2+1)^2=x^8+2x^6+3x^4+2x^2+1;=(x^4+x^2+1)^2-(x^6+x^4+x^2)-x^4=[(x^4+x^2+1)^2-x^2(x^4+x^2+1)+(x^2/2)^2]-(5/4)x^4=(x^4+x^2+1-x^2/2)^2-[(根号5/2)x^2]^2=[x^4+(根号5+1)x^2/2+1][x^4+(1-根号5)x^2+1]=[(x^4+2x^2+1)-(3+根5)x^2/2]*[(x^4+2x^2+1)-(3-根5)x^2/2]={(x^2+1)^2-[(根5-1)x/2]^2}{(x^2+1)^2-[(根5+1)x/2]^2}=[x^2+(根5-1)x/2+1][x^2-(根5-1)x/2+1][x^2+(根5+1)x/2+1][x^2-(根5+1)x/2+1].

Answer 2

原式=(x^10-1)/(x^2-1)
如果不用分式，则该式没有整系数的因式分解！

Answer 3

x的8次方+x的6次方+x的4次方+x的2次方+1
=x^6(x^2+1)+x^2(x^2+1)+1
=(x^2+1)(x^6+1)+1