Cubic Ratios

Note that we don't need to know that $p$ is a prime number to solve this question.

We have $x - p = p^{1/3} + p^{2/3}$ , then cubing both sides gives $(x - p)^3 = p + p^2 + 3p (\, \underbrace{p^{1/3} + p^{2/3}}_{=\, x-p}\, ) \quad \implies \quad x^3 + x^2 (-3p) + x(3p^2 - 3p) + (-p^3 + 2p^2 - p)=0 .$ Thus, $(a,b,c) = (-3p, 3p^2 - 3p, -p^3 + 2p^2 - p) \quad \implies \quad \dfrac{-p^3+2p^2 - p}{3p^2 - 3p} - \dfrac{3p^2 - 3p}{-3p} = 100$ Simplifying the equation above gives $p = 151 \quad\implies \quad (a,b,c) = (-453 , 67950, - 3397500)$ The answer is $(-453) + 67950 + (- 3397500) + 151 = \boxed{-3329852} .$

Let $x = \sqrt[3]{p} + \sqrt[3]{p^2} + p$ .

Then $x^3 + ax^2 + bx + c$

$= (\sqrt[3]{p} + \sqrt[3]{p^2} + p)^3 + a(\sqrt[3]{p} + \sqrt[3]{p^2} + p)^2 + b(\sqrt[3]{p} + \sqrt[3]{p^2} + p) + c$

$= (6p^2 + 3p)\sqrt[3]{p} + (3p^2 + 6p)\sqrt[3]{p^2} + p^3 + 7p^2 + p + 3ap\sqrt[3]{p} + (2ap + a)\sqrt[3]{p^2} + ap^2 + 2ap + b\sqrt[3]{p} + b\sqrt[3]{p^2} + bp + c$

$= (6p^2 + 3p + 3ap + b)\sqrt[3]{p} + (3p^2 + 6p + 2ap + a + b)\sqrt[3]{p^2} + p^3 + 7p^2 + p + ap^2 + 2ap + bp + c$

For the $\sqrt[3]{p}$ terms to cancel out, $6p^2 + 3p + 3ap + b = 0$ , and for the $\sqrt[3]{p^2}$ to cancel out, $3p^2 + 6p + 2ap + a + b = 0$ , and these two equations solve to $a = -3p$ and $b = 3p(p - 1)$ for $p \neq 1$ .

Substituting $a = -3p$ and $b = 3p(p - 1)$ back into the cubic equation, $c$ solves to $c = -p(p - 1)^2$ .

If $\cfrac{c}{b} - \cfrac{b}{a} = 100$ , then $\cfrac{-p(p - 1)^2}{3p(p - 1)} - \cfrac{3p(p - 1)}{-3p} = 100$ , which solves to $p = 151$ .

If $p = 151$ , then $a = -3p = -453$ , $b = 3p(p - 1) = 67950$ , $c = -p(p - 1)^2 = -3397500$ , and $a + b + c + p = \boxed{-3329852}$ .