Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site 
states: 

Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
  “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”
In other words, according to Example $1$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$

Should:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$
not be:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$
as everything in the left-hand side is cubed?

No! Generally speaking, one shows by induction that $\,1^r+2^r+\dots+n^r\,$ has a closed form which is a polynomial in $n$ of degree $\color{red}{r+1}$.
Examples:


*

*$1 +2 +\dots+n =\dfrac{n(n+1)}2$.

*$1^2+2^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}6$

*$1^0+2^0+\dots+n^0=\underbrace{1+1++\dots+1}_{n \ \text{times}}=n$


and the formula you posted about. What you propose hasn't the required degree, so it can't be true.
