<s>[INST] 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?
 [/INST] Well you have:
$$1^3+\ldots+n^3=\frac{n^2(n+1)^2}{4} $$
You can prove it easily by induction. Now your proposition could not work without even knowing the formula for the following reason:
$$\frac{n^3+(n+1)^3}{2}\text{ is equivalent to } n^3$$
Whereas I claim that:
$$\int_0^{n}x^3dx \leq \sum_{k=1}^nk^3\leq \int_0^{n+1}x^3dx$$
From which it follows that:
$$\sum_{k=1}^nk^3\text{ is equivalent to } \frac{n^4}{4}$$
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