Given min. polynomial of a, find min. polynomial of 1/a

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Homework Statement


Given that the minimal polynomial of a over rationals is x^4+x+8, find the minimal polynomial for 1/a over Q.


Homework Equations


I know there is a lot of work done out there for finding the min. polynomials of a^k for k>0, however I've never seen anything with a^k for k<0. I have no intuition where to start with this problem, just wondering if there are methods out there.
 
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You know that

a^4+a+8=0

An obvious thing to do is to multiply both sides by \frac{1}{a} multiple times.
 
Yea I realized this soon after posting, thanks for the help anyway.
 

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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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