How to solve for X in Cubic poynomial I've been given a Polynomial (Cubic)
$$k=\frac16x\cdot(x+1)\cdot(2x+1)$$
If $k$ is given, is there any way to solve for $x$?
 A: We solve the problem in a very special case only, where $x$ happens to be a positive integer. 
Since $\frac{x(x+1)(2x+1)}{6}\gt \frac{x^3}{3}$, we have $x^3\gt 3k$, and therefore $x\gt (3k)^{1/3}$. 
Note that $2(x+1)^3=2x^3+6x^2+6x+2$ while $x(x+1)(2x+1)=2x^3+3x^2+x$. It follows that if $x$ is a positive integer then $\frac{x(x+1)(2x+1)}{6}\lt \frac{(x+1)^3}{3}$, and therefore $x+1\gt (3k)^{1/3}$. 
It follows that $x=\lfloor (3k)^{1/3}\rfloor$. 
A: In the case where $3888 k^2-1 \gt 0$,that is to say $| k| \gt \frac{1}{36 \sqrt{3}}$ and $k>0$, the only real solution is given by $$x=\frac{1}{6} \left(\sqrt[3]{3 \sqrt{11664 k^2-3}+324
   k}+\frac{3^{2/3}}{\sqrt[3]{\sqrt{11664 k^2-3}+108 k}}-3\right)$$ This is Cardano formula applied to your equation.
Now, if $k$ is large, an asymptotic expansion of the solution is given by $$x=\sqrt[3]{3k} -\frac{1}{2}+\frac{\sqrt[3]{\frac{1}{k}}}{12
   \sqrt[3]{3}}+O\left(\left(\frac{1}{k}\right)^{4/3}\right)$$ Compare to  André Nicolas's answer and comments.
