$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$ 
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?

I have somewhat weird way to see this. Consider the system
\begin{align*} 
a + b p^{1/3} + c p^{2/3} & = 0\\
cp + a p^{1/3} + b p^{2/3} & = 0\\
bp + cp p^{1/3} + a p^{2/3} & = 0
\end{align*}
or
$$\begin{pmatrix} a & b & c \\ 
cp & a & b \\
bp & cp & a \end{pmatrix} \begin{pmatrix} 1 \\ p^{1/3} \\ p^{2/3} \end{pmatrix} =0.
$$
So the coefficient matrix has zero determinant, i.e.
$$a^3 + b(b^2-3ac)p + c^3 p^2=0.$$
Now we can proceed with infinite descent.
(The essence is until here, and below is just some calculations.)

Note that we can assume that $p$ is an integer; If $a + b (n/d)^{1/3} + c (n/d)^{2/3}=0$ then
$$ad^2 + bd\cdot  d^{2/3} n^{1/3} + c d^{4/3}n^{2/3}=0,$$
i.e.
$$ad^2 + bd\cdot   (d^2n)^{1/3} + c (d^2n)^{2/3}=0$$ so we are reduced to the integer $p$ case.
Let $q$ be a prime factor or $p$. One can assume that $q^3 \not\mid p$; in this case $q$ factor is absorbed into coefficients $b$ and $c$. Assume $(a, b, c)$ is a nontrivial integer solution. We have two cases;

*

*Case 1: Let $p = qN$ with $q\not\mid N$. Then
$$a^3 + b(b^2-3ac)qN + c^3 q^2N^2=0,$$
i.e. $a = qA$. Then
$$q^2A^3 + b(b^2-3qAc)N + c^3 qN^2=0,$$
i.e. $b = qB$. Then again
$$qA^3 + B(q B^2-3Ac)qN + c^3 N^2=0,$$
i.e. $q|c$, i.e. $c = qC$, and
$$A^3 + B(B^2-3AC)qN + C^3 q^2 N^2 = A^3 + B(B^2-3AC)p + C^3 p^2 =0.$$
Thus, if $(a, b, c)$ is an integer solution then $(a/q, b/q, c/q)$ also is an integer solution; this descent cannot be done infinitely since $a, b, c$ are finite, i.e. a contradiction.


*Case 2 : Let $p = q^2 N $ with $q\not\mid N$. Then
$$a^3 + b(b^2-3ac)q^2 N + c^3 q^4 N^2=0.$$
One can assert $a = q A$, then
$$q A^3 + b(b^2-3qAc) N + q^2c^3 N^2=0,\quad \mathbf{(**)}$$
i.e. $b = qB$. Thus
$$ A^3 + B(q B^2-3Ac)q N + q c^3  N^2=0,$$
i.e. $A$ can be divided by $q$ once again. Let $A = qA'$ to have
$$ q^2 A'^3 + B( B^2-3A'c)q N + c^3 N^2=0,$$
i.e. $c = qC$,
$$ q A'^3 + B( B^2-3qA'C) N + q^2 C^3 N^2=0 \quad \mathbf{(**)}$$
Compare two equations marked by (**);
If $(A, b, c)$ satisfies
$$q A^3 + b(b^2-3qAc) N + q^2c^3 N^2=0 $$ then we have another integer solution $(A/q, b/q, c/q)$. So, again by infinite descent, there is no such $(a, b, c)$.

This method also works for $p^{1/4}$ case.
I think I have never seen the matrix of the form
$$\begin{pmatrix} a & b & c \\ 
cp & a & b \\
bp & cp & a \end{pmatrix} $$
or its variants. Are there any reference?
