An operation similiar to division of vector? I need to solve these three equations to get $\vec x,\vec y,\vec z$ in terms of $\vec a,\vec b,\vec c$:

$$\vec x \times (\vec y \times \vec z)=\vec a\\
\vec y \times (\vec z \times \vec x)=\vec b\\
(\vec x \times \vec y)=\vec c$$


From Jacobi identity, $\vec a + \vec b =- \vec z \times (\vec x \times \vec y) =\vec c \times \vec z$
Only if I could do some division I could proceed, is there a similar operation, not division if.
Also these two can be written:
$$\vec y(\vec x.\vec z)-\vec z(\vec x.\vec y)=\vec a\\\vec z(\vec y.\vec x)-\vec x(\vec y.\vec z)=\vec b$$
This proceeds further:
$$\vec y(\vec x.\vec z)-\vec a=\vec x(\vec y.\vec z)+\vec b\\
\vec y(\vec x.\vec z)-\vec x(\vec y.\vec z)=\vec a+\vec b$$
In a way we'll again arrive at the Jacobi Identity and nowhere else.
Any suggestion to solve not only these equations but also all such vector equation?
Update:
The question is incomplete: With or without using the result obtained, obtain a solution when $|x|=|y|=|z|=\sqrt2$ and angle between each pair of $x,y,z$ is $\pi/3$. The answers are :

$$x=a\times c\\z=b+a\times c\\y=a+b+a\times c$$

 A: This is not a full answer, but too long for a comment.
First I very much doubt that there exists a general algorithm for such computations with cross products (or more general Lie algebras), because the problem is highly nonlinear in character and there are no inverses due to $a\times a=0$. But for this case..
From the Jacobi identity $x\times (y\times z)+y\times (z\times x)+z\times \ldots=0$ it follows that $a+b+z\times c=0$, so $c$ must be perpendicular to $a+b$ whenever a solution exists. 
From the conditions is follows that $x\perp c$, $y\perp c$, $x\perp a$ and $y\perp b$, which can give you the directions of $x,y$. If $a,c$ are linearly independent and $b,c$ are linearly independent, then $x$ is a multiple of $c\times a$ and $y$ a multiple of $c\times b$. The third equation then yields
$$
x\times y=\alpha\,(c\times a)\times (c\times b)=\alpha (c\cdot(a\times b))\,c
$$
for a nonzero $\alpha$ which implies that $c\cdot (a\times b)\neq 0$ if $c\neq 0$. So, we showed that if $a,c$ are independent, $b,c$ are independent and $c\neq 0$, then $a,b,c$ must be linearly independent whenever a solution exists.
Further, $y\times z\perp a$ which implies that $\{y,z,a\}$ are linearly dependent and similarly, $\{x,z,b\}$ are linearly dependent, so $z$ is in the intersection of $\langle y,a\rangle$ and $\langle x,b\rangle$, whenever these are $2$-planes. Moreover, $z\perp a+b$.
Although I don't know how to prove it, I couldn't find a counter-example for the following
Conjecture if $a,b,c$ are linearly independent and $c\perp a+b$, then a solution already exists.
(I would guess that in this case, the solution could be $x\simeq c\times a$, $y\simeq c\times b$ and $z\simeq a-\alpha b$ but I'm lazy to finish all these computations)
For the remaining cases, if $a$ resp. $b$ is a multiple of $c$ and $c\perp a+b$, then the only case is $a=\alpha c$ and $b=-\alpha c$ for some $\alpha\in\mathbb{R}$. But this is easy: let $u,v,c$ be a positively oriented orthonormal basis, then $x=u+v$, $y=-\alpha u$, $z=c$ is a solution, for example.
If $c=0$, then $x,y$ are linearly dependent, and a solution exists iff $a=-b$.
Update of the question After the update of the question, you have much stricter requirements on the solution. In fact, if you want $x,y,z$ to be all of length $\sqrt{2}$ and with pairwise angles $\pi/3$, there are only $2$ such basis up to an orthogonal orientation-preserving rotation. All equations here are equivariant with respect to orientation preserving rotations, so you may assume, WLOG, that the solution is $x=(\sqrt{2},0,0)$, $y=(\sqrt{2}/2, \sqrt{6}/2,0)$ and $z=\pm (\sqrt{3}/2, 1/2, \alpha)\beta$ where $\alpha=\sqrt{\sqrt{3}-1}$ and $\beta=\sqrt{\frac{\sqrt{2}}{\sqrt{3}}}$ or something like that. The $\pm$ includes the two possibilities; either $x,y,z$ is a positive or a negative basis. Then you can compute $a,b,c$ from the starting equations and verify whether the equations you propose $(x=a\times c, etc ...)$ are satisfied. I'm too lazy to do it. If no, then there is never a solution. If yes (for $z=+(\ldots)$ resp. $-(\ldots)$ resp. both) then there is only one $a,b,c$ up to a rotation (resp. orientation-preserving rotation) so that a solution exists. 
So you see, the problem is probably not very meaningful with all these restrictions. If fact, if you already have an answer (your constraint on $x,y,z$ are so strong that it's already almost unique) then the only thing you need to do is to verify whether this is really an answer. But it might be that there is a misunderstanding or a mistake in the assignment.
A: Given $a\in\mathbb{R}^3$ define $T_a:\mathbb{R}^3\to\mathbb{R}^3$ as $T_a(x)=a\times x$. Then, your problem is to solve the system:
$$
\begin{align}
T_x(T_y(z)) &= a \\
T_y(T_z(x)) &= b \\
T_x(y)&=c
\end{align}
$$
It's easy to see that $T$ is a linear transformation and, if $a=(a_1,a_2,a_3)$, its associated matrix is given by:
$$
T_a(z) = 
\begin{pmatrix}
0 & a_3&-a_2 \\
-a_3&0 &a_1 \\
a_2& -a_1&0 \\
\end{pmatrix}
\begin{pmatrix}
z_1 \\ z_2 \\ z_3
\end{pmatrix}.
$$
The determinant of this matrix is $\det T_a = 0$, so it's non-invertible. This means that the equation $T_a(z) = b$ for $b\neq 0$ doesn't have an unique solution (it either has an infinite number or none). This is why this problem is so hard to solve.
I don't have the general solution, but aside from Peter Franek's (very complete) answer (which unfortunately I read after writing my whole answer ;( ), I think I can help you a bit (I've managed to reduce the number of unknowns to $4$):
1) The third equation implies $x$ and $y$ lie in the orthogonal plane to $c\neq 0$. If $c=0$ they are linearly dependent.
2) By the Jacobi identity and the third equation we know that
$$
T_y(T_z(x)) = -T_x(T_y(z)) - T_z(T_x(y)) = -T_x(T_y(z)) -T_z(c) = b
$$
so, by the first equation

$$
-a -T_z(c) = b\iff T_c(z) = a+b.
$$

Write $z$ as the sum of the general solution of the homogeneous system (the space of solutions to the homogeneous system is the subspace generated by $c$) and a particular solution of the non-homogeneous system: $z = \zeta c + z_{p}$, for some $\zeta\in\mathbb{R}$. Finding the particular solution shouldn't be hard, but as we observed above this is a family of solutions. If $a=-b$ then $z_p = 0$.
3) The following two assertions hold: 
$$
-T_z(x) = T_x(z) = T_x(\zeta c) + T_x(z_p) = \zeta T_x(c) + T_x(z_p)
$$
and
$$
T_y(z) = T_y(\zeta c) + T_y(z_p) = \zeta T_y(c) + T_y(z_p).
$$ 
So, plugging this in the original equations:
$$
\begin{align}
T_y(T_z(x)) &= -\zeta T_y(T_x(c)) - T_y(T_x(z_p)) \\
&= -\zeta (x(y\cdot c)- c(x\cdot y)) - T_y(T_x(z_p)) \\
&= \zeta(x\cdot y) c - T_y(T_x(z_p)) = a.
\end{align}
$$
and 
$$
\begin{align}
T_x(T_y(z)) &= \zeta T_x(T_y(c)) + T_x(T_y(z_p)) \\
&= \zeta (y(x\cdot c) - c(x\cdot y)) + T_x(T_y(z_p)) \\
&= - \zeta(x\cdot y) c+ T_x(T_y(z_p)) = b
\end{align}
$$
Now, summing both expressions:
$$
T_x(T_y(z_p))- T_y(T_x(z_p)) = (T_xT_y- T_yT_x)(z_p) = a+b
$$
BAC-CABing once again:

$$
\begin{align}
T_x(T_y(z_p))- T_y(T_x(z_p)) &= y(x\cdot z_p) - z_p(x\cdot y) - x(y\cdot z_p) + z_p(x\cdot y) \\
&= y(x\cdot z_p) - x(y\cdot z_p) = a+b 
\end{align}
$$

I think here we could use that $x$ is orthogonal to $a$ and $y$ is orthogonal to $b$ (for example, if $c=0$ and $x$ and $y$ are linearly dependent), but ATM I can't think how. 
P.D. The las equation is maybe too good. I hope there isn't anything wrong here, I'll proof read it soon.
Edit 1: The last equation implies that $a+b\perp c$.
Edit 2: The new equations put some constraints to the vectors $a$,$b$ and $c$. Let's see some of them. Since for every $x,y\in\mathbb{R}^3$ we know that 
$$
|T_x(y)| = |x||y|\sin \theta_{xy},
$$
1') We know that $|x|=\sqrt{2} = |a||c|\sin\theta_{ac}$, so the length of at least one of these vectors is greater than $1$. 
2') $T_c(z) = T_c(b+T_a(c)) = T_c(b) + a(c\cdot c) - c(a\cdot c) = a+b$. Without loss of generality we can make $|c| = 1$, which implies
$$
T_c(b) - c(a\cdot c) = b.
$$
since $T_c(b)$ is orthogonal to $b$ and $c$, taking the dot product of this expression with $T_c(b)$:
$$
|T_c(b)| = 0
$$
i.e., $b$ must be collinear to $c$!
3')
$$
\begin{align}
T_y(x) = T_a(T_a(c)) + T_b(T_a(c)) &= a(a\cdot c) - c(a\cdot a) + a(b\cdot c) - c(a\cdot b) \\
&= a((a+b)\cdot c) - c(a\cdot(a+b)) \\ &= - c(a\cdot(a+b)) 
\end{align}
$$
Now, using the third of the original equations, $T_x(y) = (a\cdot(a+b))c=c$, so

$$
(a\cdot(a+b)) = 1
$$ 

(Check why this may not be $\sqrt{3}$). 
Let's consider now the first of the original equations.
$$
\begin{align}
T_z(y) &= T_b(a) +T_b(T_a(c)) - T_a(T_a(c)) - T_b(T_a(c))\\
&= T_b(a) - T_a(T_a(c))  \\
&= T_b(a) -a(a\cdot c) + c(a\cdot a)\\
\end{align}
$$
Then, (using the fact that $b$ and $c$ are collinear):
$$
\begin{align}
T_x(T_z(y)) &=  T_a(c)\times T_b(a) + (a\cdot c)T_a(T_a(c)) - (a\cdot a)T_c(T_a(c))\\
&= 0 + (a\cdot c)(a(a\cdot c) - c(a\cdot a)) - (a\cdot a)(a-c(a\cdot c))\\
&= ((a\cdot c) - (a\cdot a))(a\cdot c)a = a
\end{align}
$$
Then,

$$ ((a\cdot c) - (a\cdot a))(a\cdot c)=1$$

Now let's work out the second equation:
$$
\begin{align}
T_z(x) &= T_b(T_a(c)) \\
&= a(b\cdot c) - c(a\cdot b)  \\
\end{align}
$$
Then,
$$
\begin{align}
T_y(T_z(x)) &= T_b(T_a(c)) \\
&= (b\cdot c)(T_a(a)+T_b(a) - T_a(T_a(c))) - (a\cdot b)(T_a(c)+T_b(c) - T_c(T_a(c)))  \\
&= (b\cdot c)(T_b(a) - T_a(T_a(c))) - (a\cdot b)(T_a(c) - T_c(T_a(c)))\\
&= (b\cdot c)T_b(a)- (a\cdot b)T_a(c)-(b\cdot c)(a(a\cdot c) - c(a\cdot a))+(a\cdot b)a(c\cdot c) - (a\cdot b)c(a\cdot c)\\
&= (b\cdot c)T_b(a)- (a\cdot b)T_a(c)+ (c\cdot c)(a\cdot a)b -(a\cdot c)^2b \\
&= T_b(a)(b\cdot c + a\cdot c)+(a\cdot a)b -(a\cdot c)^2b \\
&= 0 + (a\cdot a)b -(a\cdot c)^2b = b
\end{align}
$$
From which,

$$
(a\cdot a) -(a\cdot c)^2 = 1
$$

If I didn't do any mistake (!!!), $(a\cdot a) = 1+(a\cdot c)^2$ and by combining the last two equations we get:
$$
((a\cdot c) - (a\cdot a))(a\cdot c) = ((a\cdot c)-1-(a\cdot c)^2)(a\cdot c) = 1
$$
So,

$$
(a\cdot c)^3-(a\cdot c)^2+(a\cdot c)+1 = 0
$$

Which only real root is $(a\cdot c)\simeq -0.54$ (it has an horrible exact form!).
Conslusion: If everything I did is right (most of it must be), given any unitary vector, $c$, it appears that we are able to give any vector $a$, with norm greater than $1$ and whose dot product with $c$ satisfies the above equation (or a similar one). Once solved, we are still missing some information to completely determine $b$, even tough we know it is collinear to $c$.
To find the norm of $b$ you could try something like:
$$
|T_z(x)| =|T_b(T_a(c))| = \sqrt{3} = |(|b|a - (a\cdot c)b)|
$$
and use the cosines law.
A: $$\newcommand{\v}[1]{{\bf \vec #1}}
\newcommand{\x}[0]{\times}
\v a\x \v c=\v x(\v y.\v y)-\v y(\v y.\v x)-\v x(\v z.\v y)+\v y(\v z.\v x)\
=2\v x-\v y-\v x+\v y=\v x\\
\fbox{$\v x=\v a\x \v c\\\v z=\v b+\v a\x\v c\\\v y=\v a+\v b+\v a\x\v c$}$$
