<s>[INST] How can i find a inverse of a polynomial in a quotient ring? I am asked to find the inverse of $\widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$. 
I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.
In previous question, we have calculated the inverse of $\widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?
Thanks a lot!!
 [/INST] extended Euc:
$$  \left(   x^{3}  + 1 \right)  $$ 
$$  \left(   x^{2}  - 2 \right)  $$ 
$$  \left(   x^{3}  + 1 \right)  =  \left(   x^{2}  - 2 \right)  \cdot \color{magenta}{  \left(   x  \right) } +  \left(  2 x  + 1 \right)  $$
$$  \left(   x^{2}  - 2 \right)  =  \left(  2 x  + 1 \right)  \cdot \color{magenta}{  \left(   \frac{ 2 x  - 1 }{ 4 }  \right) } +  \left( \frac{ -7}{4 } \right)  $$
$$  \left(  2 x  + 1 \right)  =  \left( \frac{ -7}{4 } \right)  \cdot \color{magenta}{  \left(   \frac{  - 8 x  - 4 }{ 7 }  \right) } +  \left( 0 \right)  $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{  \left(   x  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   x  \right) }{ \left( 1  \right) } $$
$$ \color{magenta}{  \left(   \frac{ 2 x  - 1 }{ 4 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{ 2 x^{2}  -  x  + 4 }{ 4 }  \right) }{ \left(   \frac{ 2 x  - 1 }{ 4 }  \right) } $$
$$ \color{magenta}{  \left(   \frac{  - 8 x  - 4 }{ 7 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{  - 4 x^{3}  - 4 }{ 7 }  \right) }{ \left(   \frac{  - 4 x^{2}  + 8 }{ 7 }  \right) } $$
$$  \left(   x^{3}  + 1 \right)  \left(   \frac{ 2 x  - 1 }{ 7 }  \right)  -  \left(   x^{2}  - 2 \right)  \left(   \frac{ 2 x^{2}  -  x  + 4 }{ 7 }  \right)  =  \left( 1  \right)  $$ 
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