unresolved partial fraction decomposition I'm having some trouble doing this  partial fraction decomposition:
$$\frac{1}{t^3-2t+1}$$
using Ruffini rule i get:
$$\frac{1}{t^3-2t+1}= \frac{1}{(t-1)(t^2+t-1)}$$
i would like to decompose the previous result into partial fraction.
I did in this way:
$$\frac{1}{(t-1)(t^2+t-1)}=\frac{A}{t-1}+\frac{B}{t^2+t-1} \leftrightarrow$$
$$\leftrightarrow t^2A+t(A+B)+(-A-B)=1$$
comparing the coefficients i get the following system of equations:\begin{cases} A=0 \\ A+B=0 \\ -A-B=1 \end{cases}
that are not true..
what am i doing wrong?

Notice, In general $$\frac{1}{(ax+b)(px^2+qx+r)}=\frac{A}{ax+b}+\frac{Bx+C}{px^2+qx+r}$$ Now, factorizing the expression, we have $$\frac{1}{(t-1)(t^2+t-1)}=\frac{A}{t-1}+\frac{Bt+C}{t^2+t-1}$$ $$\implies \frac{1}{(t-1)(t^2+t-1)}=\frac{A(t^2+t-1)+(Bt+C)(t-1)}{(t-1)(t^2+t-1)}$$ $$\implies (A+B)t^2+(A-B+C)t-(A+C)=1$$ Now, comparing the corresponding coefficients of both the sides, we get $$\begin{cases}
A+B=0\\
A-B+C=0\\
A+C=-1
\end{cases}$$ On solving the above three equations, we get $A=1, B=-1$ & $C=-2$ 
Hence, the required partial fractions are as follows $$\frac{1}{(t-1)(t^2+t-1)}=\frac{(1)}{t-1}+\frac{(-1)t+(-2)}{t^2+t-1}$$$$=\color{blue}{\frac{1}{t-1}-\frac{t+2}{t^2+t-1}}$$ 
