Dual Jacobi identity for Lie bialgebra

I'm studying Lie bialgebras: https://en.wikipedia.org/wiki/Lie_bialgebra.

I'm a bit confused about the way of writing the so called "dual Jacobi identity".

On Majid's book "a Quantum group Primer" it is written as $$\tag{1} (\delta \otimes 1 )\circ \delta (X_i) + cyclic =0,\,\,\,\,\,X_i \in g$$where $$\delta:g \to g\otimes g$$ is the skew symmetric co-commutator map, $$g$$ a Lie algebra.

My question is very simple: what is "$$+cyclic$$" in (1)? I understand the basis dependent identity but I'm unable to write down the other 2 bits in the basis independent version.. I guess one is simply: $$(1 \otimes \delta )\circ \delta (X_i)$$ but what can one rotate next? (1) needs to be applied to a single Lie algebra element $$X$$..

Thank you!

another way to write co-acobi equaton is the following: Consider $$g$$ as vector space ad $$\Lambda g$$ (the exterior algebra on $$g$$) as algebra. From a linear map $$\delta:g\to g\wedge g$$, you can extend it as a degree 1 derivation (notice $$\Lambda g$$ is super commutative and free as such, so any super-derivation is determined by its values on $$g$$. Also, since the degree is given by the tensor degree, $$\delta:g\to\Lambda^2g$$ is a degree 1 linear map). So, what is the meaning of extending by superderivation? write using Sweedler-type notation $$\delta x = x_1\wedge x_2$$ then, the extended derivation (let us call $$\delta$$ again) is given by (for $$a,b\in g$$) $$\delta(a\wedge b)=\delta( a)\wedge b - a\wedge \delta b\in\Lambda^3 g$$ Notice also (always $$a,b\in g$$) $$\delta( a)\wedge b - a\wedge \delta( b) =\delta( a)\wedge b - \delta( b)\wedge a$$ so, you can write the above formula in terms of $$\delta\otimes 1$$ and anti-symmetrization maps.
The full anti-symmetrization map (from $$g^{\otimes 3}$$ to $$\Lambda^3 g$$) has 6 terms, but if you already know some partial antisymmetry (e.g. $$\sum "a\otimes b"=\delta x\in\Lambda^2 g$$ is antisymmetric) then the six terms are actually three terms twice.
The same happens if you write Jacoby identity as $$\sum_{\sigma\in S_3}(-1)^\sigma[[x_{\sigma(1)},x_{\sigma(2)}],x_{\sigma(3)}] =2([[x_1,x_2],x_3]+[[x_2,x_3],x_1]+[[x_3,x_1]x_2])$$ just because $$[x_i,x_j]=-[x_j,x_i]$$.
As a final remark, instead of using the description of Majid's book, I prefer the definition "co-Jacoby is the condition $$\delta^2=0$$, where $$\delta$$ is the unique super-derivation on $$\Lambda g$$ with restriction to $$g$$ equal to the cobracket."