Find $n$ for a $\sum_{x=1}^n \left[(x+1)^3-x^3\right]$ $$ \sum_{x=1}^n \left[(x+1)^3-x^3\right]$$
This is my sum, I tried simplfifying and got $3x^2+3x+1$ but Im stuck on how to resolve the sum for $n$.

As you stated:
$$ \sum_{x=1}^{n} [(x+1)^3-x^3] = \sum_{x=1}^{n} 3x^2+3x+1 = 3\frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + n = $$ $$ = n(n+1)(n+2)+n = (n+1)^3-1$$
Keep in mind that:
$$ \sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}  $$
$$ \sum_{x=1}^{n} x = \frac{n(n+1)}{2}$$
