Mathematics

Resolve into partial fraction $$\displaystyle \frac{x^3-3x-2}{(x^2+x+1)(x+1)^2}$$


ANSWER

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


SOLUTION
Let $$\displaystyle \frac { x^{ 3 }-3x-2 }{ \left( x^{ 2 }+x+1 \right) \left( x+1 \right) ^{ 2 } } =\frac { A }{ \left( x+1 \right)  } +\frac { B }{ { \left( x+1 \right)  }^{ 2 } } +\frac { Cx+D }{ \left( { x }^{ 2 }+x+1 \right)  } $$
$$\Rightarrow x^{ 3 }-3x-2=A\left( x+1 \right) \left( { x }^{ 2 }+x+1 \right) +B\left( { x }^{ 2 }+x+1 \right) +\left( Cx+D \right) { \left( x+1 \right)  }^{ 2 }\\ \Rightarrow x^{ 3 }-3x-2=A\left( { x }^{ 3 }+2{ x }^{ 2 }+2x+1 \right) +B\left( { x }^{ 2 }+x+1 \right) +C\left( { x }^{ 3 }+2{ x }^{ 2 }+x \right) +D\left( { x }^{ 2 }+2x+1 \right) $$
On comparing coefficients 
$$A+C=1,2A+B+2C+D=0,2A+B+C+2D=-3,A+B+D=-2\\ \Rightarrow A=-3,B=2,C=3,D=-1$$
Hence
$$\displaystyle \frac { x^{ 3 }-3x-2 }{ \left( x^{ 2 }+x+1 \right) \left( x+1 \right) ^{ 2 } } =\frac { -3 }{ \left( x+1 \right)  } +\frac { 2 }{ { \left( x+1 \right)  }^{ 2 } } +\frac { 3x-1 }{ \left( { x }^{ 2 }+x+1 \right)  } $$
Hence, option 'C' is correct.
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