if x^3+px+r and 3x^2+p have a common factor, then p^3/27+r^2/4=0

let x= a be the common factor of both the given eqns.
then x = a must satisfy both.

.i.e

a^3 +ap + r = 0 …..............(1)

and  3a^2 + p = 0
a = + root(-p/3)  and a = -root(-p/3)...........(2)   ====> p
substitute eqn. (2) in eqn. (1)

(-p/3)^(3/2) + p*(root(-p/3)) + r  =0 
(-p/3)^(3/2) + p*(root(-p/3))  = -r
squaring both sides and after divide the whole eqn. by 4 u will get the result. 

Note that 3x²+p is the derivative of x³+px+r and hence the common root ‘a’ is at least a double root of the cubic. If a is a triple root then since sum of roots is 0, a=0=p=r and the statement is true.

 

If a is a double root, then the third root is -2a.

Hence product of roots = -r = -2a³…..........................1

Also, from the quadratic we have a² = -p/3..................2 

From the above two eqns the given relation easily follows.

Let x-a is the common factor.

Then putting X=a we get a^2= -p/3 from 2nd equn

Then 

Putting X=a in 1st eqn we get

a^3+pa+r=0

a(a^2+p)+r=0

a(-p/3+p)+r=0

a= - 3r/2p

 

Now put this value of a to the 1st eqn.

r^2/4 + p^3/27=0