Solve the equation $10x^3-6x^2-12x-8=0$ Solve the equation $10x^3-6x^2-12x-8=0$
I was doing this question but got stuck and hence looked at the solution, which went as follows $(x+2)^3=x^3+6x^2+12x+8$ and hence the equation is written as $11x^3=(x+2)^3$ from which we have that $\frac{x+2}{x}=\sqrt[3]{11}$ and hence $x=\frac{2}{\sqrt[3]{11}-1}$.
Could you please explain to me how they thought of using the fact that $(x+2)^3=x^3+6x^2+12x+8$ and if there is a more intuitive approach, could you please explain it to me?
I imagine the first thing is to try to factor.  That will fail as this has not rational roots but with $8 = 2^3$ as the last term of a third degree polynomial they think in terms of $(ax \pm 2)^3 = a^3 \pm 6a^2 + 12 a \pm 3$.  That fails as there is no such possible $a$ but the coefficients of $6,12, 8$ are just too convenient to ignore.
$(x\pm 2)^3 = x^3 \pm 6x^2 + 12x \pm 8$ and $P(x) = 10x^3 - 6x^2 -12x -8$.  So if we subtracted those then we'd wipe out every coefficient and something useful may come of it.  We'd get, if nothing else, a shorter equation.
So $P(x) + (x+2)^3 = (10x^3 - 6x^2 -12x -8) + (x^3 + 6x + 12x + 8) = 11x^3$.
Well that's a clean expression $P(x) + (x+2)^3 = 11x^3$ but what can we do with it?

I'll be honest my first thought is not what the book did, but to set $P(x) = 11x^3 - (x+2)^3$ and factor knowing $m^3 - n^3 = (m-n)(m^2 + mn + n)$.  But then I'd realize $11$ doesn't factor nicely.  But... I suppose I must be honest.  I didn't solve this before reading the books answer so if I hadn't read the solution .... I'd probably plow ahead.


$11x^3 - (x+2)^3 = (\sqrt[3]{11}x - (x+2))[(\sqrt[3]{11}^2x^2 + \sqrt[3]{11}x(x+2) +(x+2)^2]$ so either $\sqrt[3]{11}x - (x+2)=0$ and $x=\frac 2{\sqrt[3]{11}-1}$ (the book's solution)


or $(\sqrt[3]{11}^2x^2 + \sqrt[3]{11}x(x+2) +(x+2)^2=(\sqrt[3]{11}^2+\sqrt[3]{11} + 1)x^2  +(2\sqrt[3]{11}+4)x + 4=0$


And by quadratic equation: $x = \frac{-(2\sqrt[3]{11}+4) \pm \sqrt{(2\sqrt[3]{11}+4)^2 - 16(\sqrt[3]{11}^2+\sqrt[3]{11} + 1)}}{2(\sqrt[3]{11}^2+\sqrt[3]{11} + 1)}$.  But that has a negative determinate so no solution.

But the book made a simple observation that would have saved me a lot of time and headache
Well, we are attempting to solve $P(x) = 0$ so we need to solve $(x+2)^3 = 11x^3$ and now we are doing what the text did.
Maybe I'd do it differently $x+2 = \sqrt[3]{11} x$ so $x-\sqrt[3]{11}x = -2$ so $x = \frac {-2}{1-\sqrt[3]{11}} = \frac 2{\sqrt[3]{11}-1}$.
If $x\in \mathbb{R}$, we have divide both side by $2$: $$5x^{3}-3x^{2}-6x-4=0$$
Now, we can see that elimiting the quadratic term by substituting $\boxed{y=x-\frac{1}{5}}$, we have $$-4-6\left(y+\frac{1}{5}\right)-3\left(y+\frac{1}{5}\right)^{2}+5\left(y+\frac{1}{5}\right)^{3}=0$$
Expand out terms of the LHS, we have $$5y^{3}-\frac{33}{5}y-\frac{132}{25}=0 \iff y^{3}-\frac{33}{25}y-\frac{132}{125}=0$$
Now, let $\boxed{y=z+\frac{r}{z}}$ where $r$ is a constant, we have
\begin{eqnarray}
-\frac{132}{125}-\frac{33}{25}\left(z+\frac{r}{z}\right)+\left(z+\frac{r}{z}\right)^{3}&amp;=&amp;0\\
 z^{3}\left[-\frac{132}{125}-\frac{33}{25}\left(z+\frac{r}{z}\right)+\left(z+\frac{r}{z}\right)^{3} \right]&amp;=&amp;0
\end{eqnarray}
So, we obtain $$z^{6}+z^{4}\left(3r-\frac{33}{25}\right)-\frac{132}{125}z^{3}+z^{2}\left(3r-\frac{33}{25}r\right)+r^{3}=0$$
Now, let $\boxed{r=\frac{11}{25}}$ and $t=z^{3}$, we have $$t^{2}-\frac{132}{125}t+\frac{1331}{15 625}=0 \implies t=\frac{121}{125}$$
So, $z^{3}=\frac{121}{125} \implies z=\frac{1}{5}(-11)^{2/3}$Finally, we have $$\boxed{x=\frac{1}{5}\left(\sqrt[3]{11}+\sqrt[3]{(11)^{2}} \right)+\frac{1}{5}}$$
and complex solution is $$\boxed{x=\frac{1}{5}-\frac{1}{10}\sqrt[3]{11}\left(1-i\sqrt{3}\right)-\frac{1}{10}\sqrt[3]{(11)^{2}}\left(1+i\sqrt{3}\right)}.$$