Answer:
See below
Step-by-step explanation:
Let the 2 roots be A and A^2.
Then A^3 = c/a and A + A^2 = -b/a.
Using the identity a^3 b^3 = (a + b)^3 - 3ab^2 - 3a^2b:-
A^3 + (A^2)^3 = ( A + A^2)^3 - 3 A.A^4 - 3 A^2. A^2
= (A + A^2)^3 - 3A*3( A + A^2)
Substituting:-
c/a + c^2/a^2 = (-b/a)^3 - 3 (c/a)(-b/a)
Multiply through by a^3:-
a^2c + ac^2 = -b^3 + 3abc
Factoring:-
ac(a + c) = 3abc - b^3
This is not the formula required in the question but let's see if the formula in the question reduces to this. If it does we have completed the proof.
a(c - b)^3 = a(c^3 - 3c^2b + 3cb^2 - b^3) = ac^3 - 3ac^2b + 3acb^2 - ab^3
c(a - b)^3 = c(a^3 - 3a^2b + 3ab^2 - b^3) = ca^3 - 3a^2bc + 3acb^2 - cb^3.
These are equal so we have
ca^3 - 3a^2bc + 3acb^2 - cb^3 = ac^3 - 3abc^2 + 3acb^2 - ab^3
The 3acb^2 cancel out so we have:-
a^3c - ac^3 = 3a^2bc - 3abc^2 + b^3c - ab^3
ac(a^2 - c^2) = 3abc( a - c) + b^3(c - a)
ac(a + c)(a -c) = 3abc(a - c) - b^3 (a - c)
Divide through by (a - c):-
ac(a + c) = 3abc - b^3 , which is the result we got earlier.
This completes the proof.
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