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Your input: perform the partial fraction decomposition of $$$\frac{2 x^{2} - 1}{x^{6} + 1}$$$

Factor the denominator: $$$\frac{2 x^{2} - 1}{x^{6} + 1}=\frac{2 x^{2} - 1}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}$$$

The form of the partial fraction decomposition is

$$\frac{2 x^{2} - 1}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}=\frac{A x + B}{x^{2} + 1}+\frac{C x + D}{x^{2} + \sqrt{3} x + 1}+\frac{E x + F}{x^{2} - \sqrt{3} x + 1}$$

Write the right-hand side as a single fraction:

$$\frac{2 x^{2} - 1}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}=\frac{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(C x + D\right) + \left(x^{2} + 1\right) \left(x^{2} + \sqrt{3} x + 1\right) \left(E x + F\right) + \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right) \left(A x + B\right)}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}$$

The denominators are equal, so we require the equality of the numerators:

$$2 x^{2} - 1=\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(C x + D\right) + \left(x^{2} + 1\right) \left(x^{2} + \sqrt{3} x + 1\right) \left(E x + F\right) + \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right) \left(A x + B\right)$$

Expand the right-hand side:

$$2 x^{2} - 1=x^{5} A + x^{5} C + x^{5} E + x^{4} B - \sqrt{3} x^{4} C + x^{4} D + \sqrt{3} x^{4} E + x^{4} F - x^{3} A + 2 x^{3} C - \sqrt{3} x^{3} D + 2 x^{3} E + \sqrt{3} x^{3} F - x^{2} B - \sqrt{3} x^{2} C + 2 x^{2} D + \sqrt{3} x^{2} E + 2 x^{2} F + x A + x C - \sqrt{3} x D + x E + \sqrt{3} x F + B + D + F$$

Collect up the like terms:

$$2 x^{2} - 1=x^{5} \left(A + C + E\right) + x^{4} \left(B - \sqrt{3} C + D + \sqrt{3} E + F\right) + x^{3} \left(- A + 2 C - \sqrt{3} D + 2 E + \sqrt{3} F\right) + x^{2} \left(- B - \sqrt{3} C + 2 D + \sqrt{3} E + 2 F\right) + x \left(A + C - \sqrt{3} D + E + \sqrt{3} F\right) + B + D + F$$

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} A + C + E = 0\\B - \sqrt{3} C + D + \sqrt{3} E + F = 0\\- A + 2 C - \sqrt{3} D + 2 E + \sqrt{3} F = 0\\- B - \sqrt{3} C + 2 D + \sqrt{3} E + 2 F = 2\\A + C - \sqrt{3} D + E + \sqrt{3} F = 0\\B + D + F = -1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=0$$$, $$$B=-1$$$, $$$C=- \frac{\sqrt{3}}{6}$$$, $$$D=0$$$, $$$E=\frac{\sqrt{3}}{6}$$$, $$$F=0$$$

Therefore,

$$\frac{2 x^{2} - 1}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}=\frac{-1}{x^{2} + 1}+\frac{- \frac{\sqrt{3} x}{6}}{x^{2} + \sqrt{3} x + 1}+\frac{\frac{\sqrt{3} x}{6}}{x^{2} - \sqrt{3} x + 1}$$

Answer: $$$\frac{2 x^{2} - 1}{x^{6} + 1}=\frac{-1}{x^{2} + 1}+\frac{- \frac{\sqrt{3} x}{6}}{x^{2} + \sqrt{3} x + 1}+\frac{\frac{\sqrt{3} x}{6}}{x^{2} - \sqrt{3} x + 1}$$$