Prove by induction that $1^3 + \dots + n^3 = (1 + \dots + n)^2$ I'm suppose to prove by induction:
$1^3 + \dots + n^3 = (1 + \dots + n)^2$
This is my attempt; I'm stuck on the problem of factoring dots.

 A: Your attempt looks OK as far as it goes (except for a missing superscript $2$ at one point, but that's not causing any further errors).
To prove that something is equal to $1^3+2^3+3^3+4^3+\cdots+n^3$, one must show that at each step the amount that gets added is the next cube.
So how much has to be added to $(1+2+\cdots+n)^2$ to get $(1+2+\cdots+n+[n+1])^2$?  We need to show that that is the next cube, $(n+1)^3$.  The amount that gets added is
$$
\begin{align}
& \phantom{={}} (1+2+\cdots+n+[n+1])^2 - (1+2+\cdots+n)^2 \\[6pt]
& = (A + [n+1])^2 - (A)^2 \\[6pt]
& = A^2 + 2A[n+1]+[n+1]^2 - A^2 \\[6pt]
& = 2A[n+1] + [n+1]^2.
\end{align}
$$
Now here it helps to know that
$$
A = 1 + 2 + 3 + \cdots+n = \frac{n(n+1)}{2}.
$$
Thus the difference above is
$$
2A[n+1]+[n+1]^2 = n(n+1)[n+1]+[n+1]^2
$$
and this simplifies to
$$
n^3+3n^2+3n+1
$$
and finally to
$$
(n+1)^3.
$$
Next, there's the problem of organizing that into a presentable proof by induction.
A: Hint:
$$
(n+1)^3 =(n+1)(n+1+n(n+1))=(n+1)(n+1+2(1+2+\dots+n))\\
$$
Then 
$$
(1 + \dots + n)^2+(n+1)^3=(n+1)^2+2(n+1)(1+2+\dots+n)+(1+2+\dots+n)^2\\
=(1+2+\dots+n+(n+1))^2
$$
