Math Help - Prove a^3+abc+c^3=0 if 1/(m^3) and 1/(n^3) are the roots of ax^2+bx+c=0, and m(n^2)=1

1. Prove a^3+abc+c^3=0 if 1/(m^3) and 1/(n^3) are the roots of ax^2+bx+c=0, and m(n^2)=1

Part 1 of question: Given ax^2+bx+c=0, and the roots are m and n, write the quadratic formula whose roots are 1/(m^3) and 1/(n^3).

Part 2 of question: In the above equation, given that m(n^2)=1, prove that c^3+abc+a^3=0.

Answer to Part 1 of question:
If ax^2+bx+c=0 and m and n are the roots, then (x-m)(x-n)=0, therefore -b/a=m+n and c/a=mn.
So, (x-1/(m^3))(x-1-(n^3))=0. Multiplying out of the brackets gives x^2- x(1/(m^3)+1/(n^3))+1/((mn)^3)
Substituting b/a=m+n and c/a=mn into x^2- x(1/(m^3)+1/(n^3))+1/((mn)^3 gives (c^3)(x^2)-x(b(b^2-3ac))+a^3 = 0

I don't know the answer to part 2.