Question

let a, b, c and P be rational numbers such that P is not a perfect cube if a + bp the power 1 upon 3 + Dp power 2/3 equal to zero ,then prove that a=b=c=0

Answer 1

★ QUADRATIC RESOLUTIONS ★

It's given that ,

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

Therefore , p¹/³ and p²/³ are not rational numbers. 

Therefore , p is neither 1 nor 0.

p²/³ = p¹/³ * p¹/³   and so they are unequal.

LHS = a + b p¹/³ + c p²/³ = 0      ---eq. (1) 
Given p is not a perfect cube. p is not 0 or 1.

Also p¹/³,p²/³are irrational.

Now , Multiplying eq.(1) by p^1/3 to get: 
c p + a p¹/³ + b p²/³ = 0 ---eq. (2)

eq. (1) / eq. (2)
a/cp = b/a = c/b

therefore , a²=cpb ; b²=ac ; c=b²/a

Substitute the value of c in (1) to get: 
a²+abp¹/³+b²p²/³ = 0

D = -3a²b² which is 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.