<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] You have the complete extended Euclid algorithm (which IMHO is the best way) given already in Will Jaggy's post.  
Given that you already extracted $x-3$ as GCD, here is a perhaps quicker way (which may not generalise much). We simplify a bit by noting we seek linear $a(x), b(x)$, s.t.:  
$a(x)\color{blue}{(x^2-3x+5)} + b(x) \color{blue}{(x-3)(x-2)}=1$.  Setting $x=2, 3$ we get $a(x) = \frac1{15}(-2x+9)$.  
Thus we are left with 
$\tfrac1{15}(-2x+9)\color{blue}{(x^2-3x+5)} + (px+q)\color{blue}{ (x-3)(x-2)}=1$
Comparing coeffs of the cubic term and the constant, $\implies p=\frac2{15},\; 3+6q = 1\implies q = -\frac13$.
So
$$\tfrac1{15}(-2x+9)\color{blue}{(x^2-3x+5)} + \tfrac1{15}(2x-5)\color{blue}{(x-3)(x-2)}=1$$
Multiply this by $\color{blue}{x-3}$ if you want to see the equation in form $a(x) p_1(x) + b(x) p_2(x) = \gcd(p_1, p_2)$.
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