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

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 $(1)$:
$$(1^3 + 2^3 + \cdots + n^3) + 2\times n(n+1)/2 \times (n + 1) + (n+1)^2 = \\(1^3 + 2^3 + \cdots + n^3) + n(n+1)^2 + (n+1)^2 =\\1^3 + 2^3 + \cdots + n^3 + (n + 1)^3$$
$\blacksquare$
