I have obtained the cubic formula from google, for some generic cubic function: $$ax^3+bx^2+cx+d=0,$$ it is expressed:
\begin{align*} x&=\sqrt[3]{\left(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)+\sqrt{\left(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)^2+\left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}}\\& + \sqrt[3]{\left(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)-\sqrt{\left(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)^2+\left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}} \\& -\frac{b}{3a}, \end{align*} I am actually considering when $a=C,b=-B,c=0,d=0$ $$B,C\in \mathbb{R},$$ $$x\geq 0,$$ So the relation becomes: \begin{align*} x&=\sqrt[3]{\frac{B^3}{27C^3}+\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(\frac{B^2}{9C^2}\right)^3}}\\& + \sqrt[3]{\frac{B^3}{27C^3}-\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(\frac{B^2}{9C^2}\right)^3}}\\& +\frac{B}{3C}, \end{align*}
Second Attempt at the solution
Going back to the original expression: \begin{align*} x&=\sqrt[3]{\frac{B^3}{27C^3}+\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(\frac{B^2}{9C^2}\right)^3}}\\& + \sqrt[3]{\frac{B^3}{27C^3}-\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(\frac{B^2}{9C^2}\right)^3}}\\& +\frac{B}{3C}, \end{align*} Simplifying the terms in the square root gives: \begin{align*} x&=\sqrt[3]{\frac{B^3}{27C^3}+\sqrt{\left(\frac{2B^6}{729C^6}\right)}}\\& + \sqrt[3]{\frac{B^3}{27C^3}-\sqrt{\left(\frac{2B^6}{729C^6}\right)}}\\& +\frac{B}{3C}, \end{align*} Given: $$\sqrt{\left(\frac{2B^6}{729C^6}\right)}=\pm\frac{\sqrt{2}B^3}{27C^3}$$ \begin{align*} x=\sqrt[3]{\frac{B^3}{27C^3}(1\pm\sqrt{2})} + \sqrt[3]{\frac{B^3}{27C^3}(1\mp \sqrt{2})} +\frac{B}{3C}, \end{align*}
\begin{align*} x=\frac{B}{3C}((1\pm\sqrt{2})^{1/3} +(1\mp \sqrt{2})^{1/3} +1), \end{align*} I have been told that the roots are $$x=0, x=\frac{B}{C};$$ I cant see that this final expression would reduce to $$x=\frac{B}{C},$$ or $$x=0,$$ regardless of my choice of combo with the $\pm$, any further advice on what to do would be appreciated :)
Initial Attempt (Incorrect)
I have attempted to find the roots as follows; starting by cubing the whole relation: \begin{align*} x^3&=\frac{B^3}{27C^3}+\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(\frac{B^2}{9C^2}\right)^3}\\& + \frac{B^3}{27C^3}-\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(\frac{B^2}{9C^2}\right)^3}\\& +\frac{B^3}{27C^3}, \end{align*} Grouping like terms: \begin{align*} x^3&=\frac{3B^3}{27C^3}+\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(\frac{B^2}{9C^2}\right)^3}\\& -\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(\frac{B^2}{9C^2}\right)^3}\\& , \end{align*} and now expanding brackets within the square root and putting over a common denominator we get: \begin{align*} x^3&=\frac{3B^3}{27C^3}+\sqrt{\frac{2B^6}{729C^6}}-\sqrt{\frac{2B^6}{729C^6}}\\& , \end{align*} $B$ and $C$ are either positive or negative constants but due to the even powers within the square roots, they are strictly positive and thus will return a real result. Taking these terms to give either a both positive or both negative result we would get: \begin{align*} x^3&=\frac{3B^3}{27C^3}, \end{align*} So taking the cube root we get: \begin{align*} x&=\frac{\sqrt[3]{3}}{3}\frac{B}{C}, \end{align*} Now, if these square rooted terms were different in signs we would get: \begin{align*} x^3&=\frac{3B^3}{27C^3}\pm2\sqrt{\frac{2B^6}{729C^6}} , \end{align*} or: \begin{align*} x^3&=\frac{3B^3}{27C^3}\pm2\sqrt{2}\frac{B^3}{27C^3} , \end{align*} \begin{align*} x^3&=\frac{B^3}{27C^3}(3\pm 2\sqrt{2}) \end{align*} \begin{align*} x&=\frac{B}{C}\frac{\sqrt[3]{(3\pm 2\sqrt{2})}}{3} \end{align*} I have been told that the roots are $$x=0, x=\frac{B}{C};$$ Clearly these are not the roots that I have obtained, but I cannot see what I have done wrong? Any advice on how I can successfully manipulate the cubic formula would be appreciated :)
The mistake is a problem with the signs, For $b=-B,a=C,c=0,d=0$ we get: \begin{align*} x&=\sqrt[3]{\left(\frac{B^3}{27C^3}\right)+\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(-\frac{B^2}{9a^2}\right)^3}}\\& + \sqrt[3]{\left(\frac{B^3}{27C^3}\right)-\sqrt{\left(\frac{B^3}{27C^3}\right)^2+\left(-\frac{B^2}{9a^2}\right)^3}} \\& +\frac{B}{3C}, \end{align*} The terms within the square root clearly become 0 and then: $$\sqrt[3]{\frac{B^3}{27C^3}}+\sqrt[3]{\frac{B^3}{27C^3}}+\frac{B}{3C},$$ $$x=\frac{B}{C},$$