Ok so
i)Roots of x^3+bx^2+cx+d=0 are a, a^2, a^3
Hence -b = a+a^2+a^3, c = a^3+a^4+a^5 and -d=a^6
Therefore \frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}=\frac{a^5+a^4+a^3}{a^6}=\frac{-c}{d}
ii) c = -a^2b
Therefore c^3 = -a^6b^3
Hence c^3 = db^3
Therefore, b^3d-c^3 = 0