First combine everything into a single fraction:
$\begin{align*} \frac{k(6k^2-3k-1)}{2} + \big(3(k+1)-2\big)^2&=\frac{k(6k^2-3k-1)}2+(3k+1)^2\\ &=\frac{k(6k^2-3k-1)+2(3k+1)^2}2\\ &=\frac{6k^3-3k^2-k+18k^2+12k+2}2\\ &=\frac{6k^3+15k^2+11k+2}2\;.\tag{1} \end{align*}$
At this point the easiest thing to do is to multiply out the desired expression,
$\frac{(k+1)\left(6(k+1)^2-3(k+1)-1\right)}2\;,$
and verify that it’s equal to $(1)$. If you want to keep working forward from $(1)$, however, you can. We know that $k+1$ should be a factor of the numerator, so we could simply try to divide it out, but we can easily confirm this: by inspection $k=-1$ is a zero of the numerator, so $k-(-1)=k+1$ is a factor. Ordinary polynomial long division or synthetic division quickly yields the factorization
$6k^3+15k^2+11k+2=(k+1)(6k^2+9k+2)\;.$
$6(k+1)^2=6k^2+12k+6$, so $6k^2+9k+2=6(k+1)^2-3k-4=6(k+1)^2-3(k+1)-1$, as desired.