The number of partial fractions of $\frac{1-2x}{x^3+1}$ is

$1-2x=A\left(x^2-x+1\right)+\left(Bx+C\right)\left(x+1\right)$

$A+B=0\to \left(1\right);-A+B+C=-2\to \left(2\right);A+C=1\to \left(3\right)$

From (1) and (2)

$2A-C=2\to \left(4\right)$

From (1) and (4)

$A=1;C=0;B=-1$

$\therefore \frac{1}{x+1}-\frac{x}{x^2-x+1}$

The number of partial fractions of $\frac{1-2x}{x^3+1}$ is 2