Help to understand polynomial factoring

Question

I'm following some proof, but got stuck at how the factoring works. I can follow this part:

$$\begin{align*} 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\ &= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\ \end{align*}$$

The next two steps are not so clear to me anymore:

$$\begin{align*} &= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\ &= \frac{(k+1)^2(k+2)^2}{4}.\\ \end{align*}$$

I understand that first $(k+1)^3$ was changed to have the same denominator as the main term (which is $4$). Can someone help me break down the steps how the polynomials are added then after that, the powers confuse me a bit.

Answer 1

The first step that is troubling you is just factorization by $(k+1)^2$.

Consider the expression of the numerator:

$$k^2(k+1)^2+4(k+1)^3$$

which is equal to

$$(k+1)^2\times k^2+(k+1)^2\times4(k+1)=(k+1)^2(k^2+4(k+1))=(k+1)^2(k^2+4k+4).$$

The next step needs you to remark that

$$(k+2)^2=k^2+2\times k\times 2+2^2=k^2+4k+4.$$

Answer 2

\begin{align} k^2(k+1)^2 + 4(k+1)^3&=k^2(k+1)^2 + 4(k+1)(k+1)^2\\ &=(k+1)^2[k^2 + 4(k+1)]\\ &=(k+1)^2(k^2 + 4k+4)\\ &=(k+1)^2(k+2)^2\\ \end{align}

Answer 3

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

Answer 4

Just looking at the numerators, $$k^2(k+1)^2+4(k+1)^3 =k^2(k+1)^2+4(k+1)(k+1)^2$$ $$ = (k+1)^2(k^2+4(k+1)) = (k+1)^2(k^2+4k+4)$$ $$=(k+1)^2(k+2)^2$$ For the second equality, I factored out the $(k+1)^2$ that appears in both terms; something of the form $ba+ca$ can be rewritten as $a(b+c)$. For the third equality, I just distributed the $4$, noting that $4(k+1)=4k+4$. Finally, $k^2+4k+4$ is of the form $a^2+2ab+b^2$, which can be rewritten as $(a+b)^2$; here, $a=k$ and $b=2$, giving us $(k+2)^2$.