Question

.在有理数范围内分解因式x^{12}+x^9+x^6+x^3+1.

Answer 1

x^{12}+x^9+x^6+x^3+1

=(x^3-1)(x^12+x^9+x^6+x^3+1)/(x^3-1)

=(x^15-1)/(x^3-1)

=(x^5-1)(x^10+x^5+1)/(x^3-1)

=(x-1)(x^4+x^3+x^2+x+1)(x^10+x^5+1)/(x^3-1)

=(x^4+x^3+x^2+x+1)(x^11-x^10+x^6-x^5+x-1)/(x^3-1)

其中x^11-x^10+x^6-x^5+x-1

=(x^11-x^5)-(x^10-x)+(x^6-1)

=(x^6-1)(x^5+1)-x(x^9-1)

=(x^3-1)(x^3+1)(x^5+1)-x(x^3-1)(x^6+x^3+1)

=(x^3-1)[(x^8+x^5+x^3+1)-(x^7+x^4+x)]

=(x^3-1)(x^8-x^7+x^5-x^4+x^3-x+1)

∴原式=(x^4+x^3+x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)