我B PART 唔識做呀>^

1.
Let the 2 roots be m and n. So m + n = -p and mn = q.
(m + n)^3 = m^3 + 3m^2n + 3mn^2 + n^3 = m^3 + n^3 + 3mn(m + n).
That is (-p)^3 = m^3 + n^3 + 3q(-p)
So m^3 + n^3 = 3pq - p^3.
(m - n^2)(n - m^2) = mn - m^3 - n^3 + (mn)^2
= q - (m^3 + n^3) + q^2
= q - ( 3pq - p^3) + q^2
= q - 3pq + p^3 + q^2..................(1)
(b) For one root to be the square of the other, that means m = n^2 or (m - n^2) = 0. That is to say (1) = 0. So
q - 3pq + p^3 + q^2 = 0 or
p^3 - 3pq + q^2 + q = 0.

a) α+β=-p
αβ=q

(α+β)^2=α^2+β^2+2αβ
α^2+β^2=(α+β)^2-2αβ=p^2-2q

α^3+β^3=(α+β)(α^2-αβ+β^2)
=(-p)(p^2-2q-q)
=-p^3+3pq

(α-β^2)(β-α^2)=αβ-α^3-β^3+(αβ)^2
=αβ-(α^3+β^3)+(αβ)^2
=q-(-p^3+3pq)+q^2
p^3-3pq+q+q^2