\(\frac{x^2+1}{x^3-1}=\frac{a}{x-1}+\frac{bx+c}{x^2+x+1}=\frac{a\left(x^2+x+1\right)+\left(bx+c\right)\left(x-1\right)}{x^3-1}=\frac{\left(a+b\right)x^2+\left(a-b+c\right)x+\left(a-c\right)}{x^3-1}\)
a+b =1
a-b+c =0
a-c =1
=> b+c=0 => b =- c
=>a =2b=-2c
=>3b=1 => b =1/3
Vậy a =2/3
b=1/3
c =-1/3