$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

You can expand 
$$\left(\frac {n(n+1)}2\right)^2=\frac 14n^2(n^2+2n+1)=\frac 14n^4+\frac 12n^3+\frac 14n^2$$
and read off $A=\frac 14,B=\frac 12,C=\frac 14,D=0,E=0$
You can also solve simultaneous equations
$$1^3=1=A1^4+B^3+C1^2+D1+E=A+B+C+D+E\\
1^3+2^3=9=A2^4+B2^3+C2^2+D2+E=16A+8B+4C+2D+E$$
and three more similar.  You should get the same result as above.
