<s>[INST] Greatest Common Divisor of two Polynomials. Find the $\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$ over $\mathbb{Q}[x]$.  Then find two polynomials $a(x),b(x) \in \mathbb{Q}[x]$ such that, $$a(x)(x^3-6x^2+14x-15) + b(x)(x^3-8x^2+21x-18)=\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$$  
I have managed to find,
$$x^3-6x^2+14x-15=(x-3)(x^2-3x+5)$$
$$x^3-8x^2+21x-18=(x-3)(x-3)(x-2)$$
Now since $x^2-3x+5$ is irreducible over $\mathbb{Q}[x]$ and so the greatest common divisor is $(x-3)$.  Now to find $a(x)$ and $b(x)$ I have no clue how to do that.  I have looked online and it seems there is extended euclidean algorithm for polynomials but I haven't formally learned it in my class yet, so I was wondering if there is another efficient way to find these polynomials. Any help is appreciated, thanks!
 [/INST] This is just the Extended Euclidean Algorithm. Instead of back-substitution, I have always preferred to write the construction steps in the style of continued fractions. Furthermore, I have always depended on the kindness of strangers.
$$  \left(   x^{3}  - 6 x^{2}  + 14 x  - 15 \right)  $$ 
$$  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  $$ 
$$  \left(   x^{3}  - 6 x^{2}  + 14 x  - 15 \right)  =  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  \cdot \color{magenta}{  \left( 1  \right) } +  \left(  2 x^{2}  - 7 x  + 3 \right)  $$ 
 $$  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  =  \left(  2 x^{2}  - 7 x  + 3 \right)  \cdot \color{magenta}{  \left(   \frac{ 2 x  - 9 }{ 4 }  \right) } +  \left(   \frac{ 15 x  - 45 }{ 4 }  \right)  $$ 
 $$  \left(  2 x^{2}  - 7 x  + 3 \right)  =  \left(   \frac{ 15 x  - 45 }{ 4 }  \right)  \cdot \color{magenta}{  \left(   \frac{ 8 x  - 4 }{ 15 }  \right) } +  \left( 0 \right)  $$ 
 $$ \frac{ 0}{1} $$ 
 $$ \frac{ 1}{0} $$ 
 $$ \color{magenta}{  \left( 1  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left( 1  \right) }{ \left( 1  \right) } $$ 
 $$ \color{magenta}{  \left(   \frac{ 2 x  - 9 }{ 4 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{ 2 x  - 5 }{ 4 }  \right) }{ \left(   \frac{ 2 x  - 9 }{ 4 }  \right) } $$ 
 $$ \color{magenta}{  \left(   \frac{ 8 x  - 4 }{ 15 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{ 4 x^{2}  - 12 x  + 20 }{ 15 }  \right) }{ \left(   \frac{ 4 x^{2}  - 20 x  + 24 }{ 15 }  \right) } $$ 
 $$  \left(   x^{2}  - 3 x  + 5 \right)  \left(   \frac{ 2 x  - 9 }{ 15 }  \right)  -  \left(   x^{2}  - 5 x  + 6 \right)  \left(   \frac{ 2 x  - 5 }{ 15 }  \right)  =  \left( -1  \right)  $$ 
 $$  \left(   x^{3}  - 6 x^{2}  + 14 x  - 15 \right)  =  \left(   x^{2}  - 3 x  + 5 \right)  \cdot \color{magenta}{  \left(   x  - 3 \right) } +  \left( 0 \right)  $$ 
 $$  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  =  \left(   x^{2}  - 5 x  + 6 \right)  \cdot \color{magenta}{  \left(   x  - 3 \right) } +  \left( 0 \right)  $$ 
 $$  \mbox{GCD} =   \color{magenta}{  \left(   x  - 3 \right) }   $$ 
 $$  \left(   x^{3}  - 6 x^{2}  + 14 x  - 15 \right)  \left(   \frac{ 2 x  - 9 }{ 15 }  \right)  -  \left(   x^{3}  - 8 x^{2}  + 21 x  - 18 \right)  \left(   \frac{ 2 x  - 5 }{ 15 }  \right)  =  \left(   -  x  + 3 \right)  $$ 
............
</s>