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.


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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.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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