Question

Let a,b,c,p be rational numbers such that p is not a perfect cube. If a+bp 1 /3 +cp 1 /3 =0 then prove that a=b=c=0 ?

Answer 1

    a, b, c and p are rational.  p is not a perfect cube.
 so p¹/³ and p²/³ are not rational numbers. Then p is not 1 or 0.
p²/³ = p¹/³ * p¹/³   and so they are not equal.

         LHS = a + b p¹/³ + c p²/³ = 0       --- (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.

Answer 2

Answer: See the attachment