The coefficient of x^3 in the expansion of fr | vedclass

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(A) Given the expression $f(x) = \frac{x^4+1}{(x^2+1)(x-1)}$.First,perform polynomial division or algebraic manipulation.Note that $x^4+1 = (x^4-1) +

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(A) Given the expression $f(x) = \frac{x^4+1}{(x^2+1)(x-1)}$.First,perform polynomial division or algebraic manipulation.Note that $x^4+1 = (x^4-1) + 2 = (x^2-1)(x^2+1) + 2$.So,$\frac{x^4+1}{(x^2+1)(x-1)} = \frac{(x^2-1)(x^2+1) + 2}{(x^2+1)(x-1)} = \frac{(x-1)(x+1)(x^2+1) + 2}{(x^2+1)(x-1)} = (x+1) + \frac{2}{(x^2+1)(x-1)}$.We know that $\frac{1}{x-1} = -(1+x+x^2+x^3+...)$ and $\frac{1}{x^2+1} = 1-x^2+x^4-x^6+...$.Thus,$\frac{2}{(x^2+1)(x-1)} = 2(1-x^2+x^4-...) 	imes -(1+x+x^2+x^3+...)$.$= -2(1+x+x^2+x^3-x^2-x^3-x^4-x^5+x^4+x^5+x^6+x^7-...)$.$= -2(1+x+0x^2+0x^3+0x^4+0x^5+x^6+...)$.The coefficient of $x^3$ in the expansion of $\frac{2}{(x^2+1)(x-1)}$ is $0$.Since the term $(x+1)$ does not contain $x^3$,the coefficient of $x^3$ in the entire expression is $0$.

The coefficient of $x^3$ in the expansion of $\frac{x^4+1}{(x^2+1)(x-1)}$ when it is expressed in terms of positive integral powers of $x$,is

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