f.4 a.maths (quadratic equation)

(a^5) - (b^5)=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)
=((a+b)^2-2ab)^(1/2))*(a^4+b^4+ab(a^2+b^2)+(ab)^2)
=((a+b)^2-2ab)^(1/2))*((a^2+b^2)^2-2a^2b^2+ab((a+b)^2-2ab)+(ab)^2)
=((a+b)^2-2ab)^(1/2))*(((a+b)^2-2ab)^2-(ab)^2+ab((a+b)^2-2ab))
ab=q
a+b=-p
(a^5) - (b^5)=((p^2-2q)^0.5)*((p^2-2q)^2-q^2+q(p^2-2q))
=((p^2-2q)^0.5)*(p^4-4p^2q+4q^2-q^2+qp^2-2q^2)
=((p^2-2q)^0.5)*(p^4-3(p^2)q+q^2)

2006-10-23 23:54:56 補充：
Oh! sorry the second line should be =((a b)^2-4ab)^(1/2))*(a^4 b^4 ab(a^2 b^2) (ab)^2)Therefore the answer should be =((p^2-4q)^0.5)*(p^4-3(p^2)q q^2)