How to find $\sum n^3$ if $\sum n^2$ is given 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?

Note that $n(n+1)(2n+1)=2n(n+1)(n+1/2)$. Letting $m=n+1/2$, we get:
$$2780\cdot 3 = (m-1/2)(m+1/2)m = m^3-m/4$$ So we want $m$ to be just over $\sqrt[3]{3\cdot 2780}$. That should give you a candidate for $n$. You should check that this $n$ works.
Note: The $m$ trick is a bit of magic, but we know that $n(n+1)(2n+1)$ grows "like" $2n^3$, so even if you don't find the $m$ trick, you can guess it will be near $\sqrt[3]{3\cdot 2780}$. The $m$ trick makes it clear that if such $n$ exists, there has to be only a single value.
