Showing that a matrix is invertible by factorisation Let $P_5$ be the vector space of polynomials of degree $\leq$ 5 over $Q$.
 $P_5 = {a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5: a_i \in Q}$ and let $D: P_5 \longrightarrow P_5$ be the linear map $D(\alpha) = \frac{d \alpha}{dx}$. By factorising the expression $D^6 - Id$ show that $D^4+D^2+Id: P_5 \longrightarrow P_5$ is invertible and write down its inverse.
 I am not quite sure how I can show that $D^4+D^2+Id$ is invertible but I tried factorising the given expression in the following way:
$$ \begin{align*} D^6 - Id &= D^6 - Id^6\\ &= (D-Id) (D^5 + D^4Id + D^3Id^2 + D^2Id^3 + DId^4 + Id^5)\\ &= (D-Id)(D^5 + D^4 + D^3 + D^2 + D + Id)\\ &= (D-Id) (D^3(D^2+D+Id) + D^2 + D+Id)\\ &= (D-Id) (D+Id) (D^4+D^2+Id)\\ &= (D^2 - Id) (D^4 + D^2 + Id)\end{align*} $$
 and now I am not sure what to do next
Thank you in advance!

In a ring $R$, $u \in R$ is invertible if $u v = v u = 1$, for some $v \in R$.  
Here $R$ is the ring (under composition) of $\mathbb{Q}$-linear operators on $P_5$, $u = D^4 + D^2$, and $1_R = Id$.  Can you see what $D^6$ does to $P_5$?  If so, you will see that you have found the $v$.  
