2013 AIME I #5: Funky Real Root The real root of the equation $8x^3-3x^2-3x-1=0$ can be written in the form $\frac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$.
I noticed that there was a $3x^2-3x-1$ present and I decided to add $(x+1)^3$ to get rid of this. When I did this, I ended up with $x=1/(9^{1/3}-1)$. I am not sure how to rationalize this crazy mess.
 A: What I will do for most cubic equations:
Step 1:

If the equation $ax^3+bx^2+cx+d=0$ has $b=0$ then skip this step. Otherwise, let $y=x+\dfrac{b}{3a}$. 

In this case:
$$8x^3-3x^2-3x-1=0$$
Let $y=x+\dfrac{b}{3a}=x-\dfrac{1}{8}\Rightarrow x=y+\dfrac{1}{8}$.
\begin{equation}\begin{aligned}
8x^3-3x^2-3x-1=0 &\Leftrightarrow 8\left(y+\dfrac{1}{8}\right)^3-3\left(y+\dfrac{1}{8}\right)^2-3\left(y+\dfrac{1}{8}\right)-1=0 \\
&\Leftrightarrow 8\left(y^3+\dfrac{3}{8}y^2+\dfrac{3}{64}y+\dfrac{1}{512}\right) -3\left(y^2+\dfrac{1}{4}y+\dfrac{1}{64}\right)-3y-\dfrac{3}{8}-1=0\\
&\Leftrightarrow 8y^3-\dfrac{27}{8}y-\dfrac{45}{32}=0\\
\end{aligned}\end{equation}
The last equation now has $b=0$.
Step 2:

After having the last equation in the form $ax^3+cx+d=0$, write it again as $x^3+\dfrac{c}{a}x+\dfrac{d}{a}=0$ (if $a\ne 1$) and find real numbers $z,t$ that satisfy ${\begin{cases}z^3+t^3=\dfrac{d}{a}\\zt=\dfrac{-c}{3a}\end{cases}}$

This is why:
$8y^3-\dfrac{27}{8}y-\dfrac{45}{32}=0$
$\Leftrightarrow y^3-\dfrac{27}{64}y-\dfrac{45}{256}=0$
If you can find $z,t\in\mathbb{R}$ so that ${\begin{cases}z^3+t^3=\dfrac{-45}{256}\\zt=\dfrac{27}{192}\end{cases}}\Leftrightarrow {\begin{cases}z^3+t^3=\dfrac{-45}{256}\\z^3t^3=\dfrac{729}{262144}\end{cases}}\Leftrightarrow {\begin{cases}z=\sqrt[3]{\dfrac{-81}{512}}\\t=\sqrt[3]{\dfrac{-9}{512}}\end{cases}}\Leftrightarrow {\begin{cases}z=\dfrac{\sqrt[3]{-81}}{8}\\t=\dfrac{\sqrt[3]{-9}}{8}\end{cases}}$  then
$y^3-3yzt+z^3+t^3=0$
$\Leftrightarrow (y+z+t)(y^2+z^2+t^2-yz-zt-ty)=0$
$\Leftrightarrow (y+z+t)2(y^2+z^2+t^2-yz-zt-ty)=0$
$\Leftrightarrow (y+z+t)((y-z)^2+(z-t)^2+(t-y)^2)=0$
$\Leftrightarrow y=-z-t$ because you will find that $z\ne t$.
$\Rightarrow x=\dfrac{1}{8}-z-t=\dfrac{1}{8}-\dfrac{\sqrt[3]{-81}}{8}-\dfrac{\sqrt[3]{-9}}{8}=\dfrac{-\sqrt[3]{-81}-\sqrt[3]{-9}+1}{8}=\dfrac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}.$
