# The roots {x_i}, i=1,2,3,...,6 of x^6+ax^3+b=0 are such that every |x_i|=1. How do you prove that, if b^2-a^2>=1, a^2-3<=b^2<=a^2+5?. Otherwise, b^2-5<=a^2<=b^2+3 ?

Nov 5, 2016

Instead, the answer is $\left\{\left(a , b\right)\right\} = \left\{\left(\pm 2 , 1\right) \left(0 , \pm 1\right)\right\}$ and the corresponding equations are ${\left({x}^{3} \pm 1\right)}^{2} = 0 \mathmr{and} {x}^{6} \pm 1 = 0.$.

#### Explanation:

The good answer from Cesereo R enabled me to modify

my earlier version, to make my answer alright.

The form $x = r {e}^{i \theta}$ could represent both real and complex

roots. In the case of real roots x, r=|x|., Agreed! Let us proceed.

In this form, with r = 1, the equation splits into two equations,

$\cos 6 \theta + a \cos 3 \theta + b = 0$ ...(1)

and

$\sin 6 \theta + a \sin 3 \theta = 0$... (2)

To be at ease, choose (3) first and use $\sin 6 \theta = 2 \sin 3 \theta \cos 3 \theta$. It gives

$\sin 3 \theta \left(2 \cos 3 \theta + a\right) = 0$, with solutions

$\sin 3 \theta = 0 \to \theta = \frac{k}{3} \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$ ...(3)

and

$\cos 3 \theta = - \frac{a}{2} \to \theta = \left(\frac{1}{3}\right) \left(2 k \pi \pm {\cos}^{- 1} \left(- \frac{a}{2}\right)\right)$,

with k as before. ... (4)

Here, $| \cos 3 \theta | = | - \frac{a}{2} | \le 1 \to a \in \left[- 2 , 2\right]$ ... (5)

(3) reduces (1) to

$1 \pm a + b = 0$ ... (6)

Using $\cos 6 \theta = 2 {\cos}^{2} 3 \theta - 1$, (4) reduces (1) to

$2 {\left(- \frac{a}{2}\right)}^{2} - 1 - {a}^{2} / 2 + b = 0 \to b = 1$... (7)

Now, from (6), $a = \pm 2$

So, (a, b) values are (+-2, 1)..

The corresponding equations are ${\left({x}^{3} \pm 1\right)}^{2} = 0 \mathmr{and} \left({x}^{6} + 1\right) = 0$

Yet, this is not wholly tallying with Cesareo"s set of values for (a, ). I think that I have to review my answer again. Considering (4) and (6) together, upon setting a = 0, b = -1. Easy to verify that $\left(a , b\right) = \left(0 , - 1\right)$is a solution and the corresponding equation is ${x}^{6} - 1 = 0$, with two real roots $\pm 1$. Here, $6 \theta = \left(4 k - 1\right) \pi \mathmr{and} \cos 6 \theta = - 1$, and so, (6) becomes b = 1, when a = 0 also. You are 100% right, Cesareo. Thank you.

Note: This is yet another proposition, However, I would recall and make a statement on how I had set the inequalities in the present question, as early as possible.

Unfortunately, my scribbling on this matter had gone to the dust bin. If this answer is right but not that, I $r e g r e t$ for the same. I have to change the question for this answer. I think fast but don't type, in sync with thinking. Bugs get easily embedded in my thoughts.

I expect Neuroscientists to endorse my explanation, for the entry of bugs in our hard work..

.

Nov 5, 2016

See below.

#### Explanation:

Supposing that $\left\{a , b\right\} \in \mathbb{R}$ we have that $b = \pm 1$
because $b = \Pi {x}_{i}$. Now making $y = {x}^{3}$ we have

${y}^{2} + a y \pm 1 = 0$ and solving for $y$

$y = - \left(\frac{a}{2}\right) \pm \sqrt{{\left(\frac{a}{2}\right)}^{2} - \left(\pm 1\right)}$ but

$\left\mid y \right\mid = \left\mid - \left(\frac{a}{2}\right) \pm \sqrt{{\left(\frac{a}{2}\right)}^{2} - \left(\pm 1\right)} \right\mid = 1$

Solving for $a$ we have $a = \left\{0 , - 2 , 2\right\}$

The equation ${x}^{6} + a {x}^{3} + b = 0$ is equivalent to one of the possibilities

${x}^{6} + {a}_{0} {x}^{3} + {b}_{0} = 0$

with

${a}_{0} = \left\{- 2 , 0 , 2\right\}$
${b}_{0} = \left\{- 1 , 1\right\}$