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!

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 $a(x) p_1(x) + b(x) p_2(x) = \gcd(p_1, p_2)$.
