Proof by induction of series summation

Question

I understand the core principles of how to prove by induction and how series summations work. However I am struggling to rearrange the equation during the final (induction step).

Prove by induction for all positive integers n,

$$\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2$$

After both proving for $n=1$ and assuming it holds true for $n=k$:

$$\sum_{r=1}^{k+1} r^3 = \frac{1}{4}k^2(k+1)^2+(k+1)^3$$

However I am unsure of how to proceed from here, the textbook says that the next step is to rearrange to give:

$$\sum_{r=1}^{k+1} r^3 = \frac{1}{4}(k+1)^2(k^2+4(k+1))$$

However I don't understand how they did this, can someone please clarify what they have done or suggest an alternative method to rearrange this equation to prove that the statement holds true for $k+1$ to give:

$$\sum_{r=1}^{k+1} r^3 = \frac{1}{4}(k+1)^2((k+1)+1)^2$$

Answer 1

They factored out the $1/4 (k+1)^2$. So, from the first part of that sum, they got a $k^2$, and second part, they got a $4 (k+1)$. Just multiply it through if you want to see why it holds.

Answer 2

$$\frac{1}{4}k^2(k+1)^2+(k+1)^3=\frac{1}{4}k^2(k+1)^2+(k+1)(k+1)^2=(k+1)^2\left(\frac14k^2+k+1\right)=(k+1)^2\left(\frac14k^2+\frac44(k+1)\right)=(k+1)^2\left(\frac14(k^2+4(k+1))\right)=\frac14(k+1)^2\left(k^2+4(k+1))\right)$$