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

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(C) Given the partial fraction decomposition: $\frac{2x}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}$.Since $x^3 - 1 = (x - 1)(x^2 + x + 1

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$A 
eq B = C$

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(C) Given the partial fraction decomposition: $\frac{2x}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}$.Since $x^3 - 1 = (x - 1)(x^2 + x + 1)$,we have $2x = A(x^2 + x + 1) + (Bx + C)(x - 1)$.Setting $x = 1$: $2(1) = A(1^2 + 1 + 1) + 0 \implies 2 = 3A \implies A = \frac{2}{3}$.Expanding the right side: $2x = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C$.Equating coefficients of $x^2$: $A + B = 0 \implies B = -A = -\frac{2}{3}$.Equating constant terms: $A - C = 0 \implies C = A = \frac{2}{3}$.Thus,$A = \frac{2}{3}$,$B = -\frac{2}{3}$,and $C = \frac{2}{3}$.Comparing these values,$A = C 
eq B$.

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

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