Show that $x^4-x^2+1$ is irreducible over $\mathbb{Q}$ My attempts: 
I cannot apply the Eisenstein's criteria here, because there is no prime number that divides the constant term i.e. $1$ Taking a translation of the form $x \rightarrow x+a$ does not solve this issue either.  
Next, I tried the mod tests: $\operatorname{mod}2$ doesn't work since $x^4-x^2+1=(x^2+x+1)(x^2-x+1)$, similarly in $\operatorname{mod}3$ $x^4-x^2+1=(x^2+1)^2$. Now I can go on and maybe eventually find a $\operatorname{mod}p$ that works, but that is very time consuming, specially in examinations. 
So, I'll use the rational root test. The possibilities for roots are $\pm 1$ and it is easy to see that neither is a root.
The only possibility left then are quadratic factors, say, $(x^2+ax+b)(x^2+cx+d)=x^4-x^2+1$
This gives me a set of equations $bd=1, a+c=0, b+d+ac=-1$. So either $b=d=1$, in which case $a=\pm \sqrt3 \notin \mathbb{Q}$, or $b=d=-1$, which gives $a=\pm i \notin \mathbb{Q}$. 
So such factorization is not possible and hence the given polynomial is irreducible.
Is this solution correct? Also, is there an easier way to solve this? Thank you. 

There are no rational roots, so no linear factors.
If $p(x)$ is a factor of $x^4-x^2+1$ then $p(-x)$ is, too.
If $x^4-x^2+1 = (x^2+a)(x^2+b)$ then $x^2-x+1=(x+a)(x+b)$. Show that $x^2-x+1$ is irreducible.
On the other hand, you'd have to have $x^4-x^2+1=(x^2+ax+b)(x^2-ax+b)$ where $b^2=1$ and $a\neq 0$. This means that $x^4+(2b-a^2)x^2+b^2 = x^4-x^2+1$.
So you need $2b-a^2=-1$. If $b=-1,$ then this means $a^2=-1$, and if $b=1$ then $a^2=3$.
