How do you write the partial fraction decomposition of the rational expression $\frac{2}{x^{3} + 1}$ $2 / (x^3 + 1)$ ?

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Answer 1 · 2015-12-14T04:19:11

$\frac{2}{3 (x + 1)} - \frac{2 (x - 2)}{3 (x^{2} - x + 1)}$ $2/(3(x+1))-(2(x-2))/(3(x^2-x+1))$

Factor the denominator as a sum of cubes.

$\frac{2}{(x + 1) (x^{2} - x + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} - x + 1}$ $2/((x+1)(x^2-x+1))=A/(x+1)+(Bx+C)/(x^2-x+1)$

$2 = A (x^{2} - x + 1) + (B x + C) (x + 1)$ $2=A(x^2-x+1)+(Bx+C)(x+1)$

$2 = A x^{2} - A x + A + B x^{2} + B x + C x + C$ $2=Ax^2-Ax+A+Bx^2+Bx+Cx+C$

$2 = x^{2} (A + B) + x (- A + B + C) + 1 (A + C)$ $2=x^2(A+B)+x(-A+B+C)+1(A+C)$

Determine the following system:
${(A+B=0),(-A+B+C=0),(A+C=2):}$

Solve to find that ${(A=2/3),(B=-2/3),(C=4/3):}$

Therefore,

$2/(x^3+1)=2/(3(x+1))-(2(x-2))/(3(x^2-x+1))$

How do you write the partial fraction decomposition of the rational expression 2 / (x^3 + 1)2x3+1 2 / (x^3 + 1)?