Factorize : $P(x)=x^6+x^2+1$ Factorize in $\mathbb{R}[x]$ or $\mathbb{Q}[x]$ :
$$P(x)=x^6+x^2+1$$
I know all roots of $P$ are complex and I don't know if there a factorization
But My try as following :
$$P(x)=x^6+x^2+1=(x^3)^2+1+2x^3-2x^3+x^2$$
$$=(x^3+1)^2+x^2-2x^3$$
I don't know how I complete?
For factorisation in $\mathbb{Q}[X]$:

As you note $X^6+X^2+1$ has no real roots.

Now $X^6+X^2+1=(X^3+X+1)^2$ modulo $2$, and this cubic is irreducible modulo $2$.

Hence, if the polynomial factorises over $\mathbb{Q}$ - and so over $\mathbb{Z}$ - it factorises as the product of two cubics. But every cubic has a real root, so we can't have such a factorisation.
It's root are not so difficult, because it is a reducible sextic, a bi-cubic to be precised

It's roots are

$[x_1 = -(\sqrt(2^(2/3)-(\sqrt(31)-3^(3/2))^(2/3))*i)/(2^(1/6)*3^(1/4)*(\sqrt(31)-3^(3/2))^(1/6)),x_2 = (\sqrt(2^(2/3)-(\sqrt(31)-3^(3/2))^(2/3))*i)/(2^(1/6)*3^(1/4)*(\sqrt(31)-3^(3/2))^(1/6)),x_3 = -\sqrt(\sqrt(3)*(\sqrt(31)-3^(3/2))^(2/3)*i+2^(2/3)*\sqrt(3)*i-(\sqrt(31)-3^(3/2))^(2/3)+2^(2/3))/(2^(2/3)*3^(1/4)*(\sqrt(31)-3^(3/2))^(1/6)),x_4 = \sqrt(\sqrt(3)*(\sqrt(31)-3^(3/2))^(2/3)*i+2^(2/3)*\sqrt(3)*i-(\sqrt(31)-3^(3/2))^(2/3)+2^(2/3))/(2^(2/3)*3^(1/4)*(\sqrt(31)-3^(3/2))^(1/6)),x_5 = -\sqrt((-\sqrt(3)*(\sqrt(31)-3^(3/2))^(2/3)*i)-2^(2/3)*\sqrt(3)*i-(\sqrt(31)-3^(3/2))^(2/3)+2^(2/3))/(2^(2/3)*3^(1/4)*(\sqrt(31)-3^(3/2))^(1/6)),x_6 = \sqrt((-\sqrt(3)*(\sqrt(31)-3^(3/2))^(2/3)*i)-2^(2/3)*\sqrt(3)*i-(\sqrt(31)-3^(3/2))^(2/3)+2^(2/3))/(2^(2/3)*3^(1/4)*(\sqrt(31)-3^(3/2))^(1/6))]$

So factoring the polynomial can be done in many ways
   1.  if I factor it into two cubics ----- I'll need to include a square root extension from a quartic polynomial
   2. If I factor it into a quadratic and a quartic ------ I'll need to include a cube root extension from a sextic polynomial

Polynomial decomposition is helpful here, I factored it but the results was large and  messy, so I'll post the simplest case

$x^6+x^2+1 = 0$

$x^4*y-x^2*y^2+1 = 0$

Where $y$ is the root of $y^3+y-1=0$