TITLE: How to find $\sum n^3$ if $\sum n^2$ is given
QUESTION [6 upvotes]: Problem : 
Find $\sum_{j=1}^n j^3$ if $\sum_{j=1}^n j^2 =2870$ 
Can we use the following method : 
$\sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$ = 2870..
( As sum of the square of first n natural number is $\frac{n(n+1)(2n+1)}{6}$)
But how do we proceed from here to get the result?

REPLY [2 votes]: One more way: 1) use perturbation method (see Concrete Mathematicxs of and 2) solve the equation you have for $n$.
The equation $2n^3+2n^2+n-6 \cdot 2870=0$ has only 1 real root: $n=20$.
Now use perturbation method, denote $S_n=\sum_{k=1}^{n}k^4$
 and $S^{\ast}_n=\sum_{k=1}^{n}k^3$
$$
S_n +(n+1)^4=\sum_{k=1}^{n}(k+1)^4+1=S_n+4 \sum_{k=1}^{n}k^3+6 \sum_{k=1}^{n}k^2+4\sum_{k=1}^{n}k+n+1
$$
Here $S_n$ cancel out, you have thee value for $\sum_{k=1}^{n}k^2$, from which you can also find the value for $\sum_{k=1}^nk$ (it is 210), so you can easily find the value for $S^{\ast}_n$.