<s>[INST] $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$ 
Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)


My approach:
Solving the quadratic, I get:
$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$
Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.
Case 2: If $b^2 - 4ac$ is not a perfect square, $b =  \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$
Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)
So the equation is not quadratic and $c = 0$.
$a + bp^\frac{1}{3} = 0$
$-\dfrac{a}{b} = p^\frac{1}{3}$
This is a contradiction and hence $b = 0$ and $a = 0$

Is there any other way to solve this?
 [/INST] Let $K= \mathbb{Q}(p^{\frac{1}{3}}) \cong \mathbb{Q}[x]/(x^3-p).$ We have the following:

*

*$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{1}{3}}) = 0 $, as the minimal polynomial of $p^{\frac{1}{3}}$ is  $x^3-p.$

*$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{2}{3}}) = 0 $, as the minimal polynomial of $p^{\frac{2}{3}}$ is  $x^3-p^2.$

*$\text{Tr}_{K/\mathbb{Q}} (a) = 3a$, for all $a \in \mathbb{Q}.$
Now applying the trace to your equation we get
\begin{align*}
\text{Tr}_{K/\mathbb{Q}}(a + bp^\frac{1}{3} + cp^\frac{2}{3}) &= \text{Tr}_{K/\mathbb{Q}}(a)+ b \text{Tr}_{K/\mathbb{Q}}(p^\frac{1}{3} ) + c  \text{Tr}_{K/\mathbb{Q}}(p^\frac{2}{3})\\
 &= \text{Tr}_{K/\mathbb{Q}}(a)+0+0 = 3a = 0= \text{Tr}_{K/\mathbb{Q}}(0),\\
\end{align*}
thus $a=0.$ Next multiply your equation by $p^{\frac{1}{3}}$, apply the trace, and conclude that $c=0.$ Repeat.

Edit:
As Paramanand Singh correctly points out the problem reduces to showing that $f(x)= x^3-p$ is the minimal polynomial of $p^{\frac{1}{3}},$ which I assumed. However, this follows directly from Eisenstein's Criterion and the fact that $f(p^{\frac{1}{3}})=0.$ In light of this information, the polynomial given by the OP is of degree $2$, thus must be the zero polynomial.
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