Sum of roots of $p-3 = 5p^{2/3} - 6p^{1/3}=0$? In the middle of a solution of a question, the equation $$p-3 = 5p^{2/3} - 6p^{1/3}=0$$ is given. I am required to find out the value of the sum of roots for the next step, which for an equation $ax^3+bx^2+cx+d$ is equal to $-b/a$. I tried to put the equation in this format by cubing both sides, however I get $$p^3-27-9p^2+27p = 125p^2 - 216p - 90p(5p^{2/3} - 6p^{1/3})$$ which is in terms of $p^{2/3}$, therefore I think it cannot be right. What should I do? Is there any other method, or have I missed something in this method?
If
$$p^3-27-9p^2+27p = 125p^2 - 216p - 90p(5p^{2/3} - 6p^{1/3})$$
(I haven't checked that your algebra is correct) you can substitute
$$5p^{2/3} - 6p^{1/3}=p-3$$
to get rid of the fractional powers.

Note.  The pair of equations $p-3 = 5p^{2/3} - 6p^{1/3}=0$ has no solution, I am assuming you meant the single equation $p-3 = 5p^{2/3} - 6p^{1/3}$.
Hint:  $p^{2/3} = (p^{1/3})^2$