sum of perfect cubes formula proof without induction

Question

is there any proof for the sum of cubes except induction supposition? there are some proofs using induction in below page Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction

Answer 1

Write

$$(k+1)^4 = k^4+4k^3+6k^2+4k+1$$

or

$$(k+1)^4 -k^4 = 4k^3+6k^2+4k+1.$$

If you sum both sides from $k=1$ to $n$, the left side telescopes and you have

$$(n+1)^4 -1 = 4 \sum_{k=1}^{n} k^3 + 6\sum_{k=1}^{n} k^2 + 4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1.$$ Using known formulae for the last 3 sums gives

$$(n+1)^4 -1 = 4 \sum_{k=1}^{n} k^3 + 6 \frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}+n$$

and you can solve this equation for the remaining sum.

Answer 2

hint

begin by $$(n+1)^4=n^4+4n^3+6n^2+4n+1$$

... $$(3+1)^4=3^4+4.3^3+6.3^2+4.3+1$$ $$(2+1)^4=2^4+4.2^3+6.2^2+4.2+1$$ $$(1+1)^4=1^4+4.1^3+6.1^2+4.1+1$$

and sum to get

$$(n+1)^4=1+4S_3+6S_2+4S_1+n $$ with $$S_1=1+2+...+n=\frac {n (n+1)}{2}$$ $$S_2=1^2+2^2+...+n^2=\frac {n (n+1)(2n+1)}{6} $$

Now you can find your sum $S_3$.