method to solve Cubic Equation solve the cubic eq with respect to $w$?  

$$8p^3(1-c)w^3-4p^2 w^2-pw-1=0 $$ 

I try to solve it by take the factor of constant term
This is more a comment.

Wolfy says
(feel free to edit this into MathJax):

(c - 1) p!=0, w = (-54 c^2 p^6 + 3 sqrt(3) sqrt(108 c^4 p^12 - 466 c^3 p^12 + 765 c^2 p^12 - 564 c p^12 + 157 p^12) + 117 c p^6 - 67 p^6)^(1/3)/(6 2^(2/3) (c p^3 - p^3)) - (24 c p^4 - 40 p^4)/(48 2^(1/3) (c p^3 - p^3) (-54 c^2 p^6 + 3 sqrt(3) sqrt(108 c^4 p^12 - 466 c^3 p^12 + 765 c^2 p^12 - 564 c p^12 + 157 p^12) + 117 c p^6 - 67 p^6)^(1/3)) - p^2/(6 (c p^3 - p^3))

so it's a mess as expected for a cubic.
This is more a comment.

Wolfy says
(feel free to edit this into MathJax):

(c - 1) p!=0, w = (-54 c^2 p^6 + 3 sqrt(3) sqrt(108 c^4 p^12 - 466 c^3 p^12 + 765 c^2 p^12 - 564 c p^12 + 157 p^12) + 117 c p^6 - 67 p^6)^(1/3)/(6 2^(2/3) (c p^3 - p^3)) - (24 c p^4 - 40 p^4)/(48 2^(1/3) (c p^3 - p^3) (-54 c^2 p^6 + 3 sqrt(3) sqrt(108 c^4 p^12 - 466 c^3 p^12 + 765 c^2 p^12 - 564 c p^12 + 157 p^12) + 117 c p^6 - 67 p^6)^(1/3)) - p^2/(6 (c p^3 - p^3))

so it's a mess as expected for a cubic.