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.

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    $\begingroup$ Multiply both the numerator and the denominator by $\sqrt[3]{81} + \sqrt[3]9 + 1$ and you will get... Hey indeed you are almost there ;) $\endgroup$ – Hw Chu May 31 '18 at 0:28
  • $\begingroup$ How did you know to multiply the denominator by that? $\endgroup$ – Dude156 May 31 '18 at 0:30
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    $\begingroup$ The formula $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$. $\endgroup$ – Hw Chu May 31 '18 at 0:30
  • $\begingroup$ But this is not a cube, it is 1/3 power. How can we apply sum of cubes? Thanks for the help btw $\endgroup$ – Dude156 May 31 '18 at 0:31
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    $\begingroup$ Oh, as $9 - 1 = \sqrt[3]9^3 - 1^3$, you will get a cube implicitly somewhere. $\endgroup$ – Hw Chu May 31 '18 at 0:32

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:


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:


$\Leftrightarrow y^3-\dfrac{27}{64}y-\dfrac{45}{256}=0$

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


$\Leftrightarrow (y+z+t)(y^2+z^2+t^2-yz-zt-ty)=0$

$\Leftrightarrow (y+z+t)2(y^2+z^2+t^2-yz-zt-ty)=0$

$\Leftrightarrow (y+z+t)((y-z)^2+(z-t)^2+(t-y)^2)=0$

$\Leftrightarrow y=-z-t$ because you will find that $z\ne t$.

$\Rightarrow x=\dfrac{1}{8}-z-t=\dfrac{1}{8}-\dfrac{\sqrt[3]{-81}}{8}-\dfrac{\sqrt[3]{-9}}{8}=\dfrac{-\sqrt[3]{-81}-\sqrt[3]{-9}+1}{8}=\dfrac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}.$


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