Proof by induction that $(1^3 + 2^3 + 3^3+\cdots+n^3) = (1 + 2 + \cdots + n)^2$

Question

I'm sitting with the proof in front of me, but I do not understand it.

$$A = \{n \in Z^{++} \mid (1^3 + 2^3 + 3^3+\cdots+n^3) = (1 + 2 + \cdots + n)^2\}$$

The first step of proof by induction is simple enough,to prove that $1 \in A$

$1^3 = 1^2$

The next step is where I get tripped up. So I add $n + 1$ to the right hand side

$$(1 + 2 + \cdots + n + (n + 1))^2 = (1 + 2 + \cdots + n)^2 + 2(1 + 2 + \cdots + n)(n + 1) + (n + 1)^2$$

My algebra is failing me here, because I do not understand how the equation was expanded.

Answer 1

The last step uses the identity: $$(a + b)^2 = a ^2 + 2ab + b^2$$

So: $$(\underbrace{1 + 2 + \cdots + n}_a + \underbrace{(n + 1)}_b)^2 =\\= (1 + 2 + \cdots + n)^2 + 2(1 + 2 + \cdots + n)(n+1) + (n + 1)^2\tag{1}$$

By induction hypotesis we have that $(1 + 2 + \cdots + n) ^2 = (1^3 + 2^3 + \cdots + n^3)$. We also use another identity, namely $(1 + 2 + \cdots + n) = n(n+1)/2$.

Combining those two facts and substituting into $(1)$: $$(1^3 + 2^3 + \cdots + n^3) + 2\times n(n+1)/2 \times (n + 1) + (n+1)^2 = \\(1^3 + 2^3 + \cdots + n^3) + n(n+1)^2 + (n+1)^2 =\\1^3 + 2^3 + \cdots + n^3 + (n + 1)^3$$

$\blacksquare$

Answer 2

For the inductive step:

$$(\underbrace{1+2\cdots+n}_{=A=\frac{n(n+1)}{2}}+(n+1))^2=\underbrace{A^2}_{=1^3+2^3+\cdots+n^3}+\underbrace{(n+1)^2+n(n+1)^2}_{=(n+1)^3}$$