Factorization in a proof of induction I have to prove the following:
$1+3^3+ ... + (2n+1)^3=(n+1)^2(2n^2+4n+1)$ by induction.
My try:
Base case, $n=1$:
$1+3^3=(2)^2(2\cdot1^2 + 4\cdot1 + 1)$, which is true.
By inductive hypothesis, assume $n=k$:
$1 + 3^3 + ... + (2k+1)^3=(k+1)^2 (2k^2 + 4k + 1)$
For $n=k+1$
$1 + 3^3 + ... + (2k+1)^3 + (2k+3)^3 = (k+2)^2 ( 2(k+1)^2 + 4(k+1) + 1)$
Using the inductive hypothesis we need to prove that
$(k+1)^2(2k^2+4k+1) + (2k+3)^3= (k+2)^2 ( 2(k+1)^2 + 4(k+1) + 1)$
and here I have the problem, because I don't know how to manipulate any of the sides of the equation to prove this. I tried in both sides but I can't find the way. One of my last attempts ended here:
$(k+1)^2(2k^2+4k+1) + (2k+3)^3= (k+2)^2 ( 2(k+1)(k+3) + 1 )$
Which is true by wolfram alpha.
PS: I'm aware that there are a lot of papers and information about this proof but I'm not searching another way or something like that, because I saw a lot of posts about proving this statement via induction, but all of them used the last  term as (2n-1)^3, and I need to prove it when it is (2n+1)^3, and the final expression is a little bit different. I just need help factoring my last step.
Any corrections on the inductive steps that i followed are appreciated too.

This may not be the most elegant solution, 
but $(k+1)^2(2k^2+4k+1) + (2k+3)^3= (k+2)^2 ( 2(k+1)^2 + 4(k+1) + 1)$ 
is true because both sides equal $2k^4+16k^3+47k^2+60k+28$.
