Is this 3 Qubit state Entangled? Is this 3-Qubit state entangled?
http://prntscr.com/8vsg0b
$|X\rangle=\frac{1}{\sqrt 2}~~|000\rangle+ \frac{i}{\sqrt 2}|111\rangle$
I've worked with 2-Qubit states and you can turn them into bipartite system
A state is said to be not entangled 
if $|X\rangle$ is part of the Hilbert space $A \otimes B$ and there is a way to rewrite $|X\rangle$ as $|X\rangle=|a\rangle\otimes|b\rangle$
where $\otimes$ is the tensor product, and $X$ is in Hilbert space resulting from the tensor product of the Hilbert spaces $A$ and $B$. $|a\rangle \in A \text{ and } |b\rangle \in B$
MY QUESTION is instead of this bi-partition, is it possible to do a multi-parition and use a similar process to determine if a state is entangled.
If this is not required, could someone please show me the process to determining if this state is entangled or not, thank you.
 A: It is an entagled state. We will prove it by contradiction. We will assume that your state is tensor product of 3 qubits.
Take 3 qubits $a_1|0\rangle+b_1|1\rangle,a_2|0\rangle+b_2|1\rangle,a_3|0\rangle+b_3|1\rangle$ and multyply them tensornally. You get that
$$(a_1|0\rangle+b_1|1\rangle)\otimes (a_2|0\rangle+b_2|1\rangle)\otimes (a_3|0\rangle+b_3|1\rangle)=\\a_1a_2a_3|000\rangle+a_1a_2b_3|001\rangle+a_1b_2a_3|010\rangle+a_1b_2b_3|011\rangle+\\+b_1a_2a_3|100\rangle+b_1a_2b_3|101\rangle+b_1b_2a_3|110\rangle+b_1b_2b_3|111\rangle.$$
Since we assume that above equals to $\frac{1}{\sqrt 2}~~|000\rangle+ \frac{i}{\sqrt 2}|111\rangle$ we end up with bunch of equations:
$$
\left\{ 
\begin{array}{c}
a_1a_2a_3=\frac{1}{\sqrt{2}} \\ 
a_1a_2b_3=0 \\ 
a_1b_2a_3=0\\
a_1b_2b_3=0\\
b_1a_2a_3=0\\
b_1a_2b_3=0\\
b_1b_2a_3=0\\
b_1b_2b_3=\frac{i}{\sqrt{2}}\\
\end{array}
\right. 
$$
We will use only the following fact:
$(\star)$ $abc=0$ if and only, if $a=0$ or $b=0$ or $c=0.$ It holds, because $\mathbb{C}$ is a field.
From the first equation we see that $a_1\neq 0$ and $a_2\neq 0$ and $a_3\neq 0.$ From the last we see that $b_1\neq 0$ and $b_2\neq 0$ and $b_3\neq 0.$ But the second equation tells us that $a_1a_2b_3=0.$ It is a contradiction because of $(\star).$
