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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?

Duccio Bertieri
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2 Answers2

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It should be $$\frac A{t-1}+\dfrac{Bt+C}{t^2+t-1}$$

See : Partial Fraction Decomposition

lab bhattacharjee
  • 274,582
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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 $$\frac{1}{(t-1)(t^2+t-1)}=\frac{(1)}{t-1}+\frac{(-1)t+(-2)}{t^2+t-1}$$$$=\color{blue}{\frac{1}{t-1}-\frac{t+2}{t^2+t-1}}$$

Harish Chandra Rajpoot
  • 37,450