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Let $\mathbb { R } [ x ]$ be the polynomial ring in one variable over $\mathbb { R }$ . Let $I$ be the ideal of $\mathbb { R } [ x ]$ generated by the polynomial $x ^ { 3 } - 8 .$ Consider the quotient ring $A = \mathbb { R } [ x ] / I$ . Find the number of elements $a$ of the ring $A$ satisfying $a ^ { 4 } - 1 = 0$ .

Please give me a method to solve the problem in detail.

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Here is a constructive, albeit somewhat cumbersome method:

First show that every element of $A$ can be represented by an element of the form $$ax^2+bx+c,$$ with $a,b,c\in\Bbb{R}$. Then compute the representative of $$(ax^2+bx+c)^4-1,$$ keeping in mind that $x^3-8=0$ holds in $A$. This will yield three polynomial equations in $a$, $b$ and $c$ of total degree $4$; solutions to this system of equations yield elements of $A$ satisfying the given relation.


The cumbersome work can be simplified by factoring $x^3-8$ and applying the Chinese remainder theorem. I'll leave the details to you.

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