Tìm các số thực A, B và C thoả mãn 1/x^3 + 1 = A/ x + 1 + Bx + C/ x^2 - x + 1

* Hướng dẫn giải

$\frac{1}{x^3+1}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}\iff \frac{1}{x^3+1}=\frac{A\left(x^2-x+1\right)+\left(Bx+C\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}$

$\iff \frac{1}{x^3+1}=\frac{\left(A{x}^2-Ax+A\right)+\left(B{x}^2+Bx+Cx+C\right)}{x^3+1}$

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

$\iff \left\{\begin{array}{l}A+B=0\\ -A+B+C=0\\ A+C=1\end{array}\right.\iff \left\{\begin{array}{l}A=\frac{1}{3}\\ B=-\frac{1}{3}\\ C=\frac{2}{3}\end{array}\right..$