<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] There's a typo in the first step.  @mjw caught it.  Note that you're applying the quadratic formula on $p^{1/3}$, so $$p^{1/3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}, \text{ if }c \ne 0.$$
Another missing link in your proof is your lack of argument about the claimed irrationality of $± \sqrt{b^2 - 4ac} - 2cp^{2/3}$.
As you handled the degenerate case "$c = 0$ and $b \ne 0$" well, we'll keep the assumption "$c \ne 0$ or $b = 0$" for the rest of the proof.  Also we assume that $\sqrt{b^2-4ac}$ is irrational.  Multiply $2c$ on both side of the above equality, then cube it.
\begin{align}
8c^3p =& -b^3 \pm 3b^2 \sqrt{b^2-4ac} - 3b(b^2-4ac) \pm (b^2-4ac)^{3/2} \\
=& -4b^3+12abc \pm 4(b^2-ac) \sqrt{b^2-4ac} \\
2c^3p =& -b^3 + 3abc \pm (b^2-ac) \sqrt{b^2-4ac}
\end{align}
Make $\sqrt{b^2-4ac}$ the subject of the above equality.  If $b^2-ac \ne 0$,
$$\sqrt{b^2-4ac} = \pm \frac{2c^3p + b^3 - 3abc}{b^2-ac} \in \mathbb{Q},$$
contradicting our assumption on the irrationality of $\sqrt{b^2-4ac}$ if $b^2 - ac \ne 0$.

Assume that $b^2 - ac = 0$.  Then $b^2 - 4ac = b^2 = -3b^2$, and $$p^{1/3} = \frac{(-b\pm\sqrt3|b|i)}{2c} = \begin{cases} \frac{b}{c} e^{2\pi i/3} \text{ or } \frac{b}{c} e^{4\pi i/3} \text{ if } b > 0 \\ \frac{b}{c} e^{\pi i/3} \text{ or } \frac{b}{c} e^{5\pi i/3} \text{ if } b < 0 \end{cases}.$$
In this case, if $b \ne 0$, we don't have the desired conclusion, since $p = \left(\dfrac{b}{c}\right)^3$ might not be an integer, so it's not a perfect cube.
If $b = 0$, we use the assumption $b^2 = ac$, we've $a = 0$ or $c = 0$.

*

*If $a = 0$, only the term $cp^{2/3} = 0$ is left in the original equation, but $p \ne 0$ as $p$ can't be a perfect cube, so $c = 0$.

*If $c = 0$, the original equation becomes $a = 0$. Done.

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