<s>[INST] Partial fraction decompostion Solve the partial fraction.
Starting out with...
$${x^2+1\over x^3-1}={x^2+1\over (x-1)(x^2+x+1)}$$
Then the partial fraction formula part of $\displaystyle {A\over x-1}+{Bx+C\over x^2+x+1}$.
Solving with the "shortcut" method of substitution, here is what came up.
Multiplying the partial fraction part by the problem, 
$${x^2+1\over x^3-1}={x^2+1\over (x-1)(x^2+x+1)}=
{A\over x-1}+{Bx+C\over x^2+x+1}
$$
is  
$$x^2+1=A(x^2+x+1)+B(Bx+C)(x-1)$$
Then letting
$x=1$ then $2=3A$, $~~A=\frac23$
$x=0$ then $1=A-C$, then $C=-\frac{1}3$
$x=-1$ then 
$$\eqalign{2&=A(-2(-B)+C)=A+2B-2C=\frac23+2B-2\left(-\frac13\right)=\frac43+2B\cr
2-\frac43&=2B\cr}
$$
$B=  \frac13$
Then factoring, the answer is $\displaystyle{1\over3}\left({2\over x-1}+{x-1\over x^2+x+1}\right)$.
Is there any possible quicker way, or is this the fastest?
 [/INST] Just to be clear, I think you meant that the partial fraction problem is reduced to finding the coefficients for
$x^2+1=A(x^2+x+1)+(Bx+C)(x-1)$
Solving with the method of substitution is indeed probably the quickest way. But note that the coefficient method reveals valuable information.
For example, since the $x^2$ on the LHS has coefficient 1, we expect that $A+B=1  \quad (Ax^2+Bx^2=x^2$)
An obvious substitution like you did would be at $x=1$. Then, $3A=2\Rightarrow A=2/3; B=1/3$
Another apparent coefficient is the last one. $1=A-C \Rightarrow C=-1/3$
Either methods work fine but the coefficient method is very powerful. In terms of efficiency, they will be around the same, with a slight advantage towards the coefficient method.
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