How to solve in radicals this family of equations for any degree $k$? Part I. Given any constant $a,b$, the equation in $x$,
$$\left(\frac{x+\sqrt{x^2+4a}}{2}\right)^{k}+\left(\frac{x-\sqrt{x^2+4a}}{2}\right)^{k}=b\tag1$$
is solvable in radicals for any degree $k$. The general solution is,
$$x = \frac{-a}{\beta^{1/k}}+\beta^{1/k},\quad\text{where}\;\beta = \frac{b+\sqrt{b^2+4a^k}}{2}$$
For example, expanding at $k=5$, we get the DeMoivre quintic,
$$x^5+5ax^3+5a^2x=b$$
Part II. Given any constant $a,b$, the equation in $x$,
$$\frac{\big(x+\sqrt{b}\big)^k+\big(x-\sqrt{b}\big)^k}{2}+a\frac{\big(x+\sqrt{b}\big)^k-\big(x-\sqrt{b}\big)^k}{2\sqrt{b}}=0\tag2$$
is also solvable in radicals for any $k$. For example, for $k=5$, we get,
$$x^5+5ax^4+10bx^3+10abx^2+5b^2x+ab^2=0$$

Question: What is the general solution of $(2)$?

 A: Let's rewrite your equation $\,(2)\,$ for $\;r:=\dfrac {\sqrt{b}}x\,$ as :
$$\tag{1}\dfrac{(1+r)^k-(1-r)^k}{(1+r)^k+(1-r)^k}=-\frac {\sqrt{b}}a$$
then from Ron Gordon's recent answer :
$$\tag{2}\tanh(k\;\operatorname{tanh^{-1}}r)=\frac{e^{(2 k \tanh^{-1} r)}-1}{e^{(2 k\tanh^{-1} r)}+1}=-\frac {\sqrt{b}}a$$
We may revert this to get one real solution :
$$\tag{3}\dfrac {\sqrt{b}}x=-\tanh\left(\frac 1k\tanh^{-1}\frac {\sqrt{b}}a\right)$$
Other solutions may be obtained by noticing that $\,e^{\large{(2 k \tanh^{-1} r+2\pi in)}}\,$ will give the same expansion in $(2)$ so that we have too :
 $$\tag{4}\tanh(k\;\operatorname{tanh^{-1}}r+\pi i n)=-\frac {\sqrt{b}}a$$
and the $k$ roots $x_n$ (for $n=0..k-1$) given by :
$$\tag{5}\boxed{\displaystyle x_n=-\frac 1{\sqrt{b}}\;\operatorname{cotanh}\left(\frac 1k\tanh^{-1}\frac {\sqrt{b}}a+\frac{\pi in}k\right)}$$
We may too revert the passage from $(1)$ to $(2)$ (with $k$ replaced by $\dfrac 1k$ and for $\alpha:=\dfrac {\sqrt{b}}a$ )
and rewrite $(5)$ as :
$$\tag{6}\frac {\sqrt{b}}x=-\dfrac{\left(\frac{1+\alpha}{1-\alpha}\right)^{1/k}-1}{\left(\frac{1+\alpha}{1-\alpha}\right)^{1/k}+1}$$
that is :
$$\tag{7}x=\sqrt{b}\;\dfrac{1+R^{1/k}}{1-R^{1/k}}$$
where $\;R^{1/k}\;$ is one of the $k\;$ $k$-th roots of $\;R:=\dfrac{1+\alpha}{1-\alpha}=\dfrac{a+\sqrt{b}}{a-\sqrt{b}}$.
