<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] Here's another way. If $a+bq+cq^2=0$, then $a=-(bq+cq^2)$, hence $a^2=b^2q^2+2bcq^3+c^2q^4$. So if $q^3=p$, then we have two equations:
$$\begin{align}
a+bq+cq^2&=0\\
(a^2-2bcp)-c^2pq-b^2q^2&=0
\end{align}$$
Multiplying both sides of the first equation by $b^2$ and the both side of the second equation by $cp$, and then adding the resulting equations, we have
$$(ab^2+a^2c-2bc^2p)+(b^3-c^3p)q=0$$
Now if $q$ is irrational (i.e., if $p$ is not a perfect cube) and the other variables are rational, then we must have $ab^2+a^2c-2bc^2p=b^3-c^3p=0$. But $b^3-c^3p=0$ for a non-cube $p$ implies $b=c=0$, in which case the original equation, $a+bq+cq^2=0$, implies $a=0$.
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