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 $(1)$?
$x = p(1 + (\frac qp)^{\frac 15})^5 \in \Bbb Q((\frac qp)^{\frac 15})$.
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.