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How can I use partial-fraction decomposition for this fraction? $$f(x)\equiv\dfrac{-x^{5}+2x^{4}-3x^{3}+4x^{2}-5x+6}{7(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1)}.$$

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    the denominator has no real zeros – 2017-01-15
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    you must do parfrac with complex numbers – 2017-01-15
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    @Dr.SonnhardGraubner Yes. I have to do with complex numbers. But, how do I solve it? – 2017-01-15

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Note that the denominator is a factor of $x^7+1$, the other factor being $x+1$. So the zeros of the denominator are at $x=\,$e$^{\pm\text i(2k+1)\pi}$ ($k=0,1,2$). Pairing off the complex conjugates gives$$f(x)\equiv\sum_{k=0}^2\frac{a_kx+b_k}{x^2-2x\cos\frac17(2k+1)\pi+1},$$where $a_k$ and $b_k$ are real constants to be determined. Substituting in six values of $x$ (e.g. $x=0$) will give six linear equations in the constants, enabling them to be determined.

On further thought, it's best to simplify first by multiplying numerator and denominator by $x+1$, to get$$f(x)\equiv\frac{-1}{7(x+1)}+ \frac1{x^7+1}.$$Then it's easier to calculate $f(x)$ at the substituted values.

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    Can you explain more acurately how you have found the zeros of the denominator x and the expression f(x). Can you help me to find the six linear equations in the constants. Thanks – 2017-01-15