Find closed formula without using induction for $\sum_{k=0}^n k^3$

With a telescoping sum: note that $$(k+1)^4-k^4= 4k^3+6k^2+4k+1.$$ Write this equation for $k=1, 2,\dots, n$ and add them. You'll need to know the sums $1+2+\dots +n$ and $1^2+2^2+\dots +n^2$, which are standard.

Let $S_n^{(r)}=\sum_{k=1}^n k^r$. Note that $$(k+1)^4-k^4=4k^3+6k^2+4k+1$$ and after summing for $k=1,\dots,n$, (the sum on the left is telescopic) we get $$(n+1)^4-1=4 S_n^{(3)}+6S_n^{(2)}+4S_n^{(1)}+n.$$ In a similar way, $$(k+1)^3-k^3=3k^2+3k+1\implies (n+1)^3-1=3S_n^{(2)}+3S_n^{(1)}+n$$ and $$(k+1)^2-k^2=2k+1\implies (n+1)^2-1=2S_n^{(1)}+n.$$ Now starting from the last one and going back we can obtain $S_n^{(1)}$, $S_n^{(2)}$, and finally $S_n^{(3)}$.

If you'll grant me that $f(n)=\sum_{k=0}^n k^3=an^4+bn^3+cn^2+dn+e,$

then we have this system to solve:

$$ f(0)=0=e\\ f(1)=1=a+b+c+d+e\\ f(2)=9=16a+8b+4c+2d+e\\ f(3)=36=81a+27b+9c+3d+e \\ f(4)=100=256a+64b+16c+4d+e. $$

From this it follows that $e=0$ and $$9-2\times1=7=14a+6b+2c\\ 36-3\times1=33=78a+24b+6c\\ 100-4\times1=96=252a+60b+12c$$

Now can you solve that $a=\dfrac14, b=\dfrac12, c=\dfrac14,$ and $d=0$?