Mathematics

$$\displaystyle \frac{x^{2}-x-1}{x^{3}-8}=\displaystyle \frac{A}{x-2}+\displaystyle \frac{Bx+C}{x^{2}+2x+4}\Rightarrow A+B=$$


ANSWER

$$1$$


SOLUTION

$$LHS=\dfrac {x^{2}-x-1}{x^{3}-8};  RHS=\dfrac {A}{x-2}+\dfrac {Bx+C}{x^{2}+2x+4}$$
$$=\dfrac {A(x^{2}+2x+4)}{x^{3}-8}+(Bx+C)(x-2)$$
$$x^{2}-x-1=A(x^{2}+2x+4)+(Bx+C)(x-2)$$
$$LHS=RHS$$
comparing the co-efficient of $$x^{2}$$ we get
$$A+B=1$$

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Single Correct Medium Published on 17th 09, 2020
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