How to factorise this expression so that (A step-by-step method of attaining is welcomed):
$$\frac{1}{4}k^2(k+1)^2 + (k+1)^3$$
is equal to
$$\frac{1}{4}(k+1)^2(k+2)^2$$
I have tried:
$$\frac{k^2(k+1)^2}{4}+(k+1)^3$$
How do I go from $\frac{1}{4}k^2(k+1)^2$ to $\frac{1}{4}(k+1)^2$ and $(k+1)^3$ to $(k+2)^2$ ; what are the methods for this?
However I do not know how to get onto the next steps which are:
(1) $$(k+1)^2(\frac{k^2}{4}+k+1)$$ (2)$$(k+1)^2(\frac{k^2+4k+4}{4})$$
$$\frac{1}{4}k^2(k+1)^2 + (k+1)^3$$
$$\frac{1}{4}k^2(k+1)^2 + (k+1)(k+1)^2$$
$$\left[\dfrac14k^2+(k+1)\right](k+1)^2$$
$$=\dfrac{k^2+4k+4}4(k+1)^2?$$
$$=\frac{1}{4}(k+1)^2(k+2)^2$$