$a,b,c,p$ 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
 A: First we prove that $a = 0$ or $b = 0$. Suppose for the moment that $c \neq 0$. And once this is done, we can prove that the others, say, $b = 0 = c$, hence all of them are zero. For if $a \neq 0 \neq b$, let $x = p^{1/3}$, then the equation becomes: $a + bx + cx^2 = 0 \Rightarrow (a+bx)^3 = (-cx^2)^3 \Rightarrow a^3 + 3a^2bx + 3ab^2x^2 + b^3x^3 = -c^3p^2 \Rightarrow a^3 + 3a^2bx + 3ab^2\cdot \dfrac{-a-bx}{c} + b^3p = -c^3p^2$. This is enough to show that $x \in \mathbb{Q}$, but it is not and it is quite easy to establish this fact that $x = p^{1/3} \notin \mathbb{Q}$. Thus this shows $a = 0$ or $b = 0$. Once you prove one of them is $0$, the rest follows easily.
A: This answer is a little fancier than what I think the OP is looking for, but it offers some perspective about how the question relates to polynomials, so I hope it's useful:
Let $\rho = p^{1/3}$. We assume $a+b\rho + c\rho^2 = 0$ with $a,b,c$ rational, and are trying to show $a=b=c=0$. Another way to put this is: we are given a quadratic polynomial $cx^2+bx+a$ of which $\rho$ is a root, and we are trying to show it is the zero polynomial. We will do this by considering the family (call it $I$) of all polynomials of which $\rho$ is a root.
This family of polynomials (i.e. this ideal $I$ in the polynomial ring $\mathbb{Q}[x]$) clearly includes $x^3-p$, because by definition of $\rho$, $\rho^3 = p$ i.e. $\rho^3-p=0$. I claim that since $p$ is not a rational cube, it does not contain any nonzero polynomial of degree lower than 3. I argue as follows:
If it contained a polynomial of degree 1, we would have $a\rho + b = 0$. If $a=0$ then clearly $b=0$. If $a\neq 0$, then $\rho = -b/a$, and therefore $\rho$ would be rational. At this point the problem reduces to proving the irrationality of the cube root of a rational number that is not a rational cube. This can be done e.g. by considering the prime factorizations of the numerator and denominator. At any rate, as long as we know $\rho$ is not rational, this lets us conclude that if $a\rho + b = 0$, $a=b=0$, i.e. $\rho$ is not the root of any nonzero linear polynomial.
Now what if $I$ contained a polynomial of degree $2$, say $f(x) = cx^2 + bx + a$ as in the OP? If $c=0$ we are back in the degree $1$ case and we know any such polynomial is zero. But if $c\neq 0$, then since $\rho$ is also a root of $x^3-p$, then $\rho$ is also a root of the GCD of $f(x)$ and $x^3-p$. Call it $g(x)$. (One can see this because $g(x)$ is expressible as a linear combination of $f(x)$ and $x^3-p$, for example using the extended Euclidean algorithm, and thus anything that is a root of both is also a root of $g(x)$.) Nothing nonzero but degree zero can have any roots at all; thus $g(x)$ is degree at least 1 (and at most 2 since it divides $f(x)$). But $g(x)$ can't be degree 1 by what we did above; therefore it is degree $2$ and is equal to $f(x)$ up to a scalar; i.e., $f(x)$ is a factor of $x^3-p$. But then let
$$h(x) = \frac{x^3-p}{f(x)}$$
Now $h(x)$ is degree $3-2 = 1$, so it is linear and has a rational root.  But meanwhile it is a factor of $x^3-p$, so any root of $h(x)$ is also a root of $x^3-p$ and is therefore a cube root of $p$ (just like $\rho$), and so cannot be rational for the same reason $\rho$ cannot. So this case too is impossible.
We conclude that the set of polynomials with $\rho$ as a root does not contain any nonzero polynomials of degree lower than 3.
Addendum: To relate this answer to NotALoner's: what I have argued here is basically that if there is a degree 2 polynomial with $\rho$ as a root, then there is also a degree $1$ polynomial with $\rho$ as a root, therefore it is rational, contradiction. NotALoner has explicitly calculated a degree 1 polynomial satisfied by $\rho$ given such a degree 2 polynomial, whereas this answer gives general grounds to believe it must exist.
A: 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, $a=b=c=0$
