$1^3+2^3+...+n^3= $? How I can find formula $1^3+2^3+...+n^3=$ has a formula which $(1+2+3+...+n)^2 $ or $[n(n + 1)/2]^2$ . We can verify with ınduction and I know how I can prove it. How we find what is the formula...
I tried like this
$1^3+2^3+...+n^3= An^4+ Bn^3+Cn^2+Dn+E$
after a long process i found it this way
$A=-299/54,$
$B=-4/9,$
$C=107/54,$
$D=4.$
And then I tried how can show that 
But I didn't find Maybe The value of a,b,c,d is wrong 
Can I get proof of this formula in this way or another way
if we didn't know the formula, how would we get this formula

$0=E, 1=A+B+C+D+E, 1^3+2^3=9=16A+8B+4C+2D+E,$
$ 1^3+2^3+3^3=36=81A+27B+9C+3D+E, $ 
and $1^3+2^3+3^3+4^3=100=256A+64B+16C+4D+E \implies$
$ E=0, 7=14A+6B+2C, 33=78A+24B+6C, $ and $96=252A+60B+12C \implies$
$12=36A+6B$ and $54=168A+24B\implies$
$24A=6\implies A=\dfrac14$.  Can you solve now for $B, C, $ and $D$?
