Real roots of an equation

Question

A friend gave me this equation which I have trouble finding the real roots. $$x^9+3x^6+3x^3-16x+9=0$$ One can easily see that 1 is a root then with the help of Horner's method this can be simplified. However I am looking for an elegant solution if possible not just to use a computer to do that madness calculation. I reduced it to $$x^3+1=2(2x-1)^{\frac{1}{3}}$$ I don't have any useful attempts.

EDIT: After several tries I think I found something useful. After getting to this form $x^3-2(2x-1)^{\frac{1}{3}}+1=0$ we might notice that it's similar to the equation that you provided me in the answers namely$$(x-1)(x^2 +x-1)=x^3 - 2x+1=0$$ Now in order that this to give me the solution, we must have that $-2(2x-1)^{\frac{1}{3}}=-2x\, $ If we cube both sides we return to $x^3 - 2x+1=0$ (the solutions) Is this a repeatedly loop in the equation? Can this be useful?

Answer 1

Let $f(x)=(x^3+1)/2$. Then $f$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$ with inverse $g(x)=(2x-1)^{1/3}$. As you already remarked, the original equation can be written as $$((x^3+1)=2(2x-1)^{1/3}.$$ This equation is equivalent to $f(x)=g(x)$ or $f(f(x))=x.$

Now note that if $f(x)=x$, that is $x^3-2x+1=0$ then $f(f(x))=x$ which implies that $x^3-2x+1$ divides the given polynomial. After the division, we obtain the factorization $$(x^3-2x+1)(\underbrace{(x^3+1)^2 + 2 x^4+3x^2+(x+1)^2}_{\geq 0} + 7) \tag{*}.$$ So it remains to solve $$x^3-2x+1=(x - 1) (x^2 + x - 1)=0$$ and the desired real roots are $x_1=1$, $x_2=(-1-\sqrt{5})/2$, and $x_3=(-1+\sqrt{5})/2$ .

P.S. In my first answer I used maple to find the factorization $(*)$.

Answer 2

The polynomial can be factored by considering the product of $$(x^3 + a x^2 + b x + c) \cdot (x^6 + \alpha x^5 + \beta x^4 + \gamma x^3 + \delta x^2 + \eta x + \mu).$$ With a little bit of effort it can be found that $$(x - 1) (x^2 + x - 1) (x^6 + 2 x^4 + 2 x^3 + 4 x^2 + 2 x + 9) = 0.$$ The real roots are then obtained as $x=1$ and roots from $x^2 + x -1=0$.

Answer 3

One way to attack this problem is by estimating number of real roots. For that we can use Descarte's Rule of Signs. So, for $$f(x) = x^9+3x^6+3x^3-16x+9$$ We have $2$ change of signs, hence the number of positive real roots are $0$ or $2$, but since we know $f(1)=0$, therefore we have $2$ positive real roots.

Now, $$f(-x) = -x^9+3x^6-3x^3+16x+9$$ We have $3$ change of signs, hence the number of negative real roots are $1$ or $3$

So, $f(x)$ has either $3$ or $5$ real roots

That's as far you can get without getting your hands dirty with ardous calculations involving estimation of coefficients as done by both Leucippus and Robert Z. But if you take your first case as $f(x)$ having $3$ real roots, then you essentially land at Leucippus's answer, and you're in luck as that works, in this case.