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?
 A: 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 $x = \frac {-2}{1-\sqrt[3]{11}} = \frac 2{\sqrt[3]{11}-1}$.
