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A dozen-and-a-half cubic equations to solve.

Posted 8 Months ago by chiarizio

I am wondering how to solve these and what the solutions are.
Can anyone tell me about the real zeroes of these polynomials?
(That is, the real numbers x at which each polynomial evaluates to zero?)

(x^3) - 1
(x^3) + 1

(x^3) - x - 1
(x^3) - x + 1
(x^3) + x - 1
(x^3) + x + 1

(x^3) - (x^2) - 1
(x^3) - (x^2) + 1
(x^3) + (x^2) - 1
(x^3) + (x^2) + 1

(x^3) - (x^2) - x - 1
(x^3) - (x^2) - x + 1
(x^3) - (x^2) + x - 1
(x^3) - (x^2) + x + 1
(x^3) + (x^2) - x - 1
(x^3) + (x^2) - x + 1
(x^3) + (x^2) + x - 1
(x^3) + (x^2) + x + 1

?

I hope someone answers!

There are 1 Replies


I know solutions for some of them.

x^3 - 1 is zero if x = 1
x^3 + 1 is zero if x = -1

x^3 + x^2 - x - 1
x^3 - x^2 + x - 1
x^3 - x^2 - x + 1
are all 0 if x = 1

x^3 + x^2 - x - 1
x^3 - x^2 - x + 1
x^3 + x^2 + x +1
are all 0 if x = -1

One of those is (x - 1)(x - 1)(x + 1) = (x - 1)(x^2 - 1) = x^3 - x^2 - x + 1 .
And one is (x - 1)(x + 1)(x + 1) = (x^2 - 1)(x + 1) = x^3 + x^2 - x - 1.

x^3 - x^2 + x - 1 is (x - 1)(x^2 + 1) so besides its real root 1 it has two conjugate imaginary roots i and -i.
x^3 + x^2 + x + 1 is (x + 1)(x^2 + 1) so besides its real root -1 it has two conjugate imaginary roots i and -i.

I’ll have to have more time and space to see if there are any other rational solutions.
Obviously all of them have at least one real root, but I don’t think any of the others are rational.
Can that be proven correct? Or can it be proven incorrect?

8 Months ago
chiarizio
 

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