1. **Problem statement:** Given rational numbers $a,b,c,p$ with $p$ not a perfect cube, prove that if $$a + b p^{\frac{1}{3}} + c p^{\frac{2}{3}} = 0,$$ then $a = b = c = 0$. 2. **Key idea:** The numbers $1$, $p^{\frac{1}{3}}$, and $p^{\frac{2}{3}}$ are linearly independent over the rationals if $p$ is not a perfect cube. This means no nontrivial rational linear combination of these can be zero. 3. **Assume for contradiction:** Suppose not all of $a,b,c$ are zero. Then the expression $$a + b p^{\frac{1}{3}} + c p^{\frac{2}{3}} = 0$$ with $a,b,c \in \mathbb{Q}$ and $p$ not a perfect cube. 4. **Rewrite:** Let $x = p^{\frac{1}{3}}$. Then the equation becomes $$a + b x + c x^2 = 0.$$ 5. **Since $x$ is a root of this polynomial with rational coefficients,** the minimal polynomial of $x$ over $\mathbb{Q}$ divides $a + b x + c x^2$. 6. **Minimal polynomial of $x$:** Since $x^3 = p$ and $p$ is not a perfect cube, the minimal polynomial of $x$ over $\mathbb{Q}$ is $$x^3 - p = 0,$$ which is irreducible over $\mathbb{Q}$. 7. **Degree contradiction:** The polynomial $a + b x + c x^2$ has degree at most 2, but the minimal polynomial of $x$ has degree 3. Therefore, the only polynomial of degree less than 3 that vanishes at $x$ is the zero polynomial. 8. **Conclusion:** Hence, $$a = b = c = 0.$$ This proves the linear independence of $1, p^{\frac{1}{3}}, p^{\frac{2}{3}}$ over $\mathbb{Q}$ when $p$ is not a perfect cube.