Melody

Yes that is right, what is the problem ?

55( mod 34) also equals 21

Did you miss 8 Heureka ?

x^3 + (p+x)x^2 +2px = 0

\(x^3 + (p+x)x^2 +2px = 0\\ x[x^2 + (p+x)x +2p]= 0\\ \text{x =0 is one solution}\\~\\ \text{If x is not 0 then}\\ x^2 + (p+x)x +2p=0\\ x^2 + x^2+px +2p=0\\ 2x^2 +px +2p=0\\ x=\frac{-p\pm\sqrt{p^2-4*2*2p}}{4}\\ x=\frac{-p\pm\sqrt{p^2-16p}}{4}\\~\\ so\\ x=\frac{-p\pm\sqrt{p^2-16p}}{4}\qquad or \qquad x=0\)

Thanks Rom   

Before you ask more completely new questions take some time to digest what you have already been given.

I am not actually sure if the first answer is correct or not.

I know the guest who wrote that. They were not insulted by my words.

But yes I could have been less blunt about it.

Here is the question:

You do not sound convinced LOL :)

No one likes my #6.              I was offering a real insight     

You can always cancel units. They work just like numbers, it can make difficult problems much easier.  !