Question 1046962
<pre><b>
{{{(13x+2)/((x^""-1)(x^2+x+1))}}} {{{""=""}}} {{{A/(x^""-1)}}}{{{""+""}}}{{{(Bx+C)/(x^2+x+1)}}}

Multiply through by {{{matrix(1,3,LCD,""="",(x^""-1)(x^2+x+1))}}}

{{{13x+2}}}{{{""=""}}}{{{A(x^2+x+1)+(Bx^""+C)(x^""-1)}}}

That must be true for every value of x.  So we may substitute
any number for x and it will be true.

We substitute x=1

{{{(13(1)+2)}}}{{{""=""}}}{{{A(1^2+1+1)+(B(1^"")+C)(1^""-1)}}}

{{{(13+2)}}}{{{""=""}}}{{{A(1^""+1+1)+(B^""+C)(0)}}}

{{{(15)}}}{{{""=""}}}{{{A(3)+0}}}

{{{(15)}}}{{{""=""}}}{{{3A}}}

{{{(5)}}}{{{""=""}}}{{{A}}}


We substitute x=0 and A=5


{{{(13(0)+2)}}}{{{""=""}}}{{{5(0^2+0+1)+(B(0)^""+C)(0^""-1)}}}


{{{2}}}{{{""=""}}}{{{5(1^"")+C(-1^"")}}}

{{{2)}}}{{{""=""}}}{{{5-C}}}

{{{C}}}{{{""=""}}}{{{3}}}

We substitute x=2, A=5, and C=3

{{{(13x+2)}}}{{{""=""}}}{{{A(x^2+x+1)+(Bx^""+C)(x^""-1)}}}

{{{(13(2)+2)}}}{{{""=""}}}{{{5(2^2+2+1)+(B*2^""+3)(2^""-1)}}}

{{{26+2}}}{{{""=""}}}{{{5(4^""+2+1)+(2B+3^"")(1^"")}}}

{{{28}}}{{{""=""}}}{{{5(7)+2B+3}}}

{{{28}}}{{{""=""}}}{{{35+2B+3}}}

{{{28}}}{{{""=""}}}{{{38+2B}}}

{{{-10}}}{{{""=""}}}{{{2B}}}

{{{-5}}}{{{""=""}}}{{{B}}}

Answer: Substitute A=5, B=-5, C=3

{{{(13x+2)/((x-1)(x^2+x+1))}}}{{{""=""}}}{{{5/(x^""-1)}}}{{{""+""}}}{{{(-5x+3)/(x^2+x+1)}}}

or if you like:

{{{(13x+2)/((x-1)(x^2+x+1))}}}{{{""=""}}}{{{5/(x^""-1)}}}{{{""+""}}}{{{(-(5x-3))/(x^2+x+1)}}}

{{{(13x+2)/((x-1)(x^2+x+1))}}}{{{""=""}}}{{{5/(x^""-1)}}}{{{""-""}}}{{{(5x-3)/(x^2+x+1)}}}

Edwin</pre></b>