# 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 ?#

##### 2 Answers

#### Answer:

Instead, the answer is

#### Explanation:

The good answer from Cesereo R enabled me to modify

my earlier version, to make my answer alright.

The form

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,

and

To be at ease, choose (3) first and use

and

with k as before. ... (4)

Here,

(3) reduces (1) to

Using

Now, from (6),

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

The corresponding equations are

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

The wholly complete answer is as entered in the answer box.

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

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

.

#### Answer:

See below.

#### Explanation:

Supposing that

because

Solving for

The equation

with