We will prove the relation by using the quadratic equation general solution.
How to prove that?
If
[tex]x^4 - x^2 +1 =0[/tex]
Then
[tex]x^2 = \frac{1 \pm \sqrt{(-1)^2 - 4*1*1} }{2} \\\\x^2 = \frac{1 \pm \sqrt{-3} }{2}[/tex]
Now, the other equation is:
[tex](1 + \frac{1}{x^{10}} )*x^5 = -\sqrt{3} \\\\(x^5 + \frac{1}{x^5} ) = -\sqrt{3}\\\\x^{10} + 1 = -\sqrt{3}*x^5[/tex]
Writing this as a quadratic:
[tex](x^5)^2 + \sqrt{3}*x^5 + 1 = 0[/tex]
[tex]x^5 = \frac{-\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4*1*1} }{2} \\\\x^5 = \frac{-\sqrt{3} \pm \sqrt{(-1)} }{2}[/tex]
Then we must have:
[tex]( \frac{1 \pm \sqrt{3}i }{2} )^{2.5} = \frac{-\sqrt{3} \pm \sqrt{1}i}{2}\\\\( \frac{1 \pm \sqrt{3}i }{2} )^{5/2} = \frac{-\sqrt{3} \pm \sqrt{1}i}{2}\\\\( \frac{1 \pm \sqrt{3}i }{2} )^{5} = (\frac{-\sqrt{3} \pm \sqrt{1}i}{2})^2[/tex]
Which can be checked with a calculator to be true for both signs (the + and -)
If you want to learn more about quadratic equations:
https://brainly.com/question/1214333
#SPJ1
