Question

Let a, b, c, and p be rational numbers such that p is not a perfect cube.

If a + b p^{1/3} + c p^{2/3} =0 then prove that a = b = c = 0.

Answer 1

$a+b p^\frac{1}{3}+ c p^\frac{2}{3}=0. .. ----(1)$

Given p is not a perfect cube. p is not 0 or 1. Also $p^\frac{1}{3} \: p^\frac{2}{3}$ are irrational.

Multiply (1) by p^1/3 to get:
$cp + a p^\frac{1}{3} + b p^\frac{2}{3} = 0. \: ... ... (2)\\(1) \: & \: (2) : \frac{a}{cp} = \frac{b}{a} = \frac{c}{b}\\\\a^2=cpb \: and \: b^2=ac. ... ...(3)\\c = \frac{b^2}{a}.$

Substitute the value of c in (1) to get: $a^2 + a b p^\frac{1}{3} + b^2 p^\frac{2}{3} = 0. .. Quadratic \\\\Discriminant = -3 a^2 b^2 = negative.$

So p^1/3 is imaginary. It is a contradiction as p is a rational number.  Given quadratic isn't valid.

So a = b = c = 0.
There is alternate method to solve it. See the enclosed picture.