A solvable quintic with the root $x=(\sqrt[5]{p}+\sqrt[5]{q})^5$ - what are the other roots? I derived a two parameter quintic equation with the root:

$$x=(\sqrt[5]{p}+\sqrt[5]{q})^5,~~~~~p,q \in \mathbb{Q}$$

$$\color{blue}{x^5}-5(p+q)\color{blue}{x^4}+5(2p^2-121pq+2q^2)\color{blue}{x^3}-5(2p^3+381p^2q+381pq^2+2q^3)\color{blue}{x^2}+\\ + 5(p^4-121p^3q+381p^2q^2-121pq^3+q^4)\color{blue}{x}-(p+q)^5=0 \tag{1}$$

I'm unsure what the other roots are.

My thoughts: this is really a $25$ degree equation, it has $25$ roots.
Since there are 5 fifth degree roots, it makes sence, that we have:
$$x=y^5$$
$$y_{kn}=w_{5k}\sqrt[5]{p}+w_{5n}\sqrt[5]{q},~~~~~k,n=\{1,2,3,4,5\}$$
With $w_{5k}$ - $5$th roots of unity and $\sqrt[5]{p},\sqrt[5]{q}$ - principal roots.

Is this correct?
If we have $25$ different roots of the $25$th degree equation, what about the quintic equation $(1)$?


Edit
Another possible way - we can probably write:
$$x=p(1+\sqrt[5]{q/p})^5=q(1+\sqrt[5]{p/q})^5$$
Then we have $5$ possible roots inside the bracket.
Is this the correct way to get all the roots of $(1)$?
 A: $x = p(1 + (\frac qp)^{\frac 15})^5 \in \Bbb Q((\frac qp)^{\frac 15})$.
And so the splitting field of your degree $5$ polynomial is $\Bbb Q((\frac qp)^{\frac 15},\zeta_5)$, the conjugates of $x$ are obtained by multiplying $(\frac qp)^{\frac 15}$ with a power of $\zeta_5$.
Also this is not really a degree $25$ equation.
If you naïvely look at the "conjugate" $(\zeta_5 p^\frac15 + \zeta_5 q^\frac 15)^5$, well it turns out you can factor out $\zeta_5$ and this is equal to your original $x$. So $x$ doesn't have as many distinct conjugates as you would think it should have.
A: I see what you are getting at. Since there are actually five solutions $\alpha$ to the equation,
$$\alpha^5 = p$$ 
then it makes sense that $p^{1/5}+q^{1/5} = z$ should be the root of a $25$th deg eqn, correct? This is essentially given by your equation slightly modified as,

Eq.1:
  $$z^{25}-5(p+q)z^{20}+5(2p^2-121pq+2q^2)z^{15}-5(2p^3+381p^2q+381pq^2+2q^3)z^{10}+\\ + 5(p^4-121p^3q+381p^2q^2-121pq^3+q^4)z^5-(p+q)^5=0 $$

However, since you raised your sum to a fifth power $x = \big(p^{1/5}+q^{1/5}\big)^5 = z^5$, you lost $20$ of the roots,

Eq.2:
  $$x^{5}-5(p+q)x^{4}+5(2p^2-121pq+2q^2)x^{3}-5(2p^3+381p^2q+381pq^2+2q^3)x^{2}+\\ + 5(p^4-121p^3q+381p^2q^2-121pq^3+q^4)x-(p+q)^5=0 $$

But there is another way to get a quintic out of $(1)$: one can factor it. This can be done by using appropriately chosen $p,q$, namely,
$$p=\tfrac{-b+\sqrt{b^2+4}}{2}, \quad q=\tfrac{-b-\sqrt{b^2+4}}{2}$$
and $(1)$ simplifies as,
$$z^{25} + 5 b z^{20} + 5 (125 + 2 b^2) z^{15} + 5 b (-375 + 2 b^2) z^{10} + 5 (625 + 125 b^2 + b^4) z^5 + b^5 = 0$$
This factors into a $5$-deg and $20$-deg $P(z)$, easily done by Wolfram Alpha or Mathematica,
$$(z^5+5z^3+5z+b)\, P(z) = 0$$
with the $5$-deg as the familiar DeMoivre quintic. Thus, the $25$ possible sums of,
$$z = p^{1/5}+q^{1/5} = \Big(\tfrac{-b+\sqrt{b^2+4}}{2}\Big)^{1/5} +\Big(\tfrac{-b-\sqrt{b^2+4}}{2}\Big)^{1/5}$$
solve the DeMoivre quintic, as well as a $20$th deg eqn you can see by clicking on the WA link above.
