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?
 A: 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. 
A: Hint
Since your polynomial has real coefficient, then 
$$p(X)=X^6+X^2+1$$
has roots $$\{z_1,\bar z_1, z_2,\bar z_2, z_3,\bar z_3\},$$
where $\bar z$ denotes the complex conjugate of $z$. Therefore you can write $p(X)$ as
 $$(X^2+a_1X+b_1)(X^2+a_2X+b_2)(X^2+a_3X+b_3)$$
where $$X^2+a_iX+b_i=(X-z_i)(X-\bar z_i).$$
A: Note that we can try if we have a purely imaginary root, and indeed we have this particular factorisation: $$(x^2+a)(x^4-ax^2+a^2+1)=x^6+x^2+a^3+a$$
Which fits the orignal polynomial for $\ a^3+a=1$, and this cubic has at least one real root that we can calculate by Cardan's method (I skip to the result though...)
Le set $\alpha=\sqrt[3]{108+12\sqrt{93}}\ $ then $\ a=\dfrac{\alpha}6-\dfrac 2{\alpha}$
Now we get to factorize the remaining quartic, there are formulas for this that should lead you to a factorization in $\mathbb R[x]$ in product of quadratics, but considering there are no odd powers we can try to factorize under the form:
$$(x^2+bx+c)(x^2-bx+c)=x^4+(2c-b^2)x^2+c^2$$
This gives us by identification of the coefficients (note that we are interested in one possible factorization, so I just take the positive square root each time) $\begin{cases}c=\sqrt{a^2+1}\\b=\sqrt{2c+a}\end{cases}$
And there you have it:
$$x^6+x^2+1=(x^2+a)(x^2+bx+c)(x^2-bx+c)$$
