$a,b,c,p$ are rational number and $p$ is not a perfect cube Given that $a,b,c,p$ are rational number and $p$ is not a perfect cube, if $a+bp^{1\over 3}+cp^{2\over 3}=0$ then we have to show $a=b=c=0$
I concluded that $a^3+b^3p+c^3p^2=3abcp$ but how can I go ahead? could you please help? Thanks

Clearly $p^{1/3}$ is one of the roots of the Quadratic equation: $cx^2+bx+a = 0$
Now, sum of roots = $-b/c$. Since RHS is rational and one of the terms in LHS is irrational, it is logical to assume that the other root is: $-p^{1/3}$ (so that the LHS becomes rational). The sum of roots then becomes $0$. Hence, $b = 0$.
Again, product of roots = $a/c$.
$\implies -p^{2/3} = a/c$ 
$\implies a/c + p^{2/3} = 0$
$\implies a + cp^{2/3} = 0$
Now, $a$ is rational and $cp^{2/3}$ is irrational (due to $p^{2/3}$ being irrational). This means both the terms are $0$.
Hence, $a = 0$ and $cp^{2/3} = 0$. And, $p \neq 0$ as it is not a perfect cube (but $0$ is). So, $c = 0$.
Hence, $a=b=c=0$
