If frac{2x}{x^3 - 1} = frac{A}{x - 1} + frac{ | vedclass

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(D) Given $\frac{2x}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}$.Multiplying both sides by $(x^3 - 1) = (x - 1)(x^2 + x + 1)$,we get:$2x

Vedclass · Accepted Answer

$A 
eq B 
eq C$

Vedclass · Answer

(D) Given $\frac{2x}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}$.Multiplying both sides by $(x^3 - 1) = (x - 1)(x^2 + x + 1)$,we get:$2x = A(x^2 + x + 1) + (Bx + C)(x - 1)$.For $x = 1$,$2(1) = A(1 + 1 + 1) + 0$ $\Rightarrow 3A = 2$ $\Rightarrow A = \frac{2}{3}$.Expanding the $RHS$: $2x = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C$.$2x = (A + B)x^2 + (A - B + C)x + (A - C)$.Comparing coefficients of $x^2$: $A + B = 0 \Rightarrow B = -A = -\frac{2}{3}$.Comparing constant terms: $A - C = 0 \Rightarrow C = A = \frac{2}{3}$.Thus,$A = \frac{2}{3}$,$B = -\frac{2}{3}$,and $C = \frac{2}{3}$.Since $A = C = \frac{2}{3}$ and $B = -\frac{2}{3}$,we have $A 
eq B$ and $B 
eq C$ and $A 
eq B 
eq C$ is satisfied as $A 
eq B$ and $B 
eq C$ and $A 
eq C$ is false,but the option $A 
eq B 
eq C$ is interpreted as $A 
eq B$ and $B 
eq C$ and $A 
eq C$. However,since $A=C$,the condition $A 
eq B 
eq C$ is the best fit among the choices provided.

If $\frac{2x}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}$,then

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