分解因式 (1)x^3-9x+8 (2)(x^4-4x^2+1)(x^4+3x^2+1)+10x^4

(1)
x^3-9x+8
=x^3-1-9x+9
=(x-1)(x^2+x+1)-9(x-1)
=(x-1)(x^2+x-8)
其中求解x^2+x-8=0得到两根x1,x2
则
x^3-9x+8=(x-1)(x-x1)(x-x2)

(2)
令x^4+1=A,x^2=B
则
(x^4-4x^2+1)(x^4+3x^2+1)+10x^4
=(A-4B)(A+3B)+10B^2
=A^2-AB-12B^2+10B^2
=A^2-AB-2B^2
=(A-2B)(A+B)
=(x^4+1-2x^2)(x^4+1+x^2)
其中
x^4+1-2x^2
=(x^2-1)^2
=(x-1)^2(x+1)^2
x^4+1+x^2
=x^4+2x^2+1-x^2
=(x^2+1)^2-x^2
=(x^2+1-x)(x^2+1+x)
则
(x^4-4x^2+1)(x^4+3x^2+1)+10x^4
=(x-1)^2(x+1)^2(x^2+1-x)(x^2+1+x)