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Answer» $\:\:\:\:\:\:\:\:\:\:\:\:\:\: \large\mathfrak{\dag \: Solution :- }\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ Formula :- $\sf\red { \: {<klux>X</klux>}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}$ $\sf \: Here, \: we \: are \: asked \: to \:resolve \: into \: partial \: fraction : \: \dfrac{x}{ {x}^{3} + 1}$ $\sf\: \dfrac{x}{ {x}^{3} + 1} = \sf \: \dfrac{x}{(x + 1)( {x}^{2} - x + 1) }\\\\$ $\sf \: Let \: \: \red { \dfrac{x}{(x + 1)( {x}^{2} - x + 1) } = \sf \: \dfrac{a}{x + 1} + \dfrac{bx + c}{ {x}^{2} - x + 1 } } - - (1)\\$ $\sf \: x \: = a( {x}^{2} - x + 1) + (bx + c)(x + 1) - - (2)\\$ $\sf :\implies \: - 1 = a(1 + 1 + 1) + 0\\$ $\sf :\implies \: - 1 = 3a\\$ $\sf:\implies \red {\:a \: = \: - \: \dfrac{1}{3} - - (3)}\\\\$ ❍On SUBSTITUTING 'x = - 0' in equation (2), we get :- $\sf :\implies\: 0 = a(0 - 0 + 1) + c(0 + 1)\\$ $\sf :\implies \:a + c = 0\\$ $\sf :\implies\:c = - \: a\\$ $\sf:\implies \red {\:\:c \: = \: \dfrac{1}{3} - - (4)}\\\\$ ❍On substituting 'x = 1' in equation (2), we get :- $\sf :\implies \:1 = a(1 - 1 + 1) + (b + c)(2)\\$ $\sf :\implies \:1 = a + 2b + 2c\\$ $\sf :\implies \:1 = - \: \dfrac{1}{3} + \dfrac{2}{3} + 2b\\$ $\sf :\implies \:1 = \dfrac{1}{3} + 2b\\$ $\sf :\implies \:2b = 1 - \dfrac{1}{3} \\$ $\sf :\implies \:2b = \dfrac{2}{3}\\$ $\sf:\implies \red {\:b \: = \: \dfrac{1}{3} \: - - (5)}\\$ ❍Now, substitute the VALUES.. $\sf \: \: \dfrac{x}{(x + 1)( {x}^{2} - x + 1) } = \sf \: \dfrac{ - 1}{3(x + 1)} + \dfrac{x + 1}{3( {x}^{2} - x + 1) }\\$ $\sf \: \: \purple{\dfrac{x}{ {x}^{3} + 1 } } = \sf \pink {\: \dfrac{ - 1}{3(x + 1)} + \dfrac{x + 1}{3( {x}^{2} - x + 1) }}\\\\$ ⠀⠀⠀ ━━━━━━━━━━━━━━━━━━━⠀ $\sf \green {Do \: you \: know \:?¿}$ Partial fraction decomposition is the breaking down of a rational expression into simpler parts. It is the opposite of adding rational expressions. When adding two rational expressions, there has to be a common denominator. $\begin{gathered}\boxed{\begin{array}{c\|c} \sf \pink { Term \: in \: denominator }& \sf\pink { Partial \: fraction \: decomposition }\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ax + b & \sf \displaystyle \sf \dfrac{A}{{ax + b}} \\ \\ \sf {(ax + b)}^{2} & \sf \displaystyle \sf \dfrac{{{A_1}}}{{ax + b}} + \sf\frac{{{A_2}}}{{{{\sf\left( {ax + b} \right)}^2}}} \\ \\ \sf {ax}^{2} + bx + c & \sf \displaystyle \sf \dfrac{{Ax + B}}{{a{x^2} + bx + c}} \end{array}} \\ \end{gathered}$ ⠀⠀⠀ ━━━━━━━━━━━━━━━━━━━⠀

Answer»

$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \large\mathfrak{\dag \: Solution :- }\:\:\:\:\:\:\:\:\:\:\:\:\:\:$

Formula :-

$\sf\red { \: {<klux>X</klux>}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}$

$\sf \: Here, \: we \: are \: asked \: to \:resolve \: into \: partial \: fraction : \: \dfrac{x}{ {x}^{3} + 1}$

$\sf\: \dfrac{x}{ {x}^{3} + 1} = \sf \: \dfrac{x}{(x + 1)( {x}^{2} - x + 1) }\\\\$

$\sf \: Let \: \: \red { \dfrac{x}{(x + 1)( {x}^{2} - x + 1) } = \sf \: \dfrac{a}{x + 1} + \dfrac{bx + c}{ {x}^{2} - x + 1 } } - - (1)\\$

$\sf \: x \: = a( {x}^{2} - x + 1) + (bx + c)(x + 1) - - (2)\\$

$\sf :\implies \: - 1 = a(1 + 1 + 1) + 0\\$

$\sf :\implies \: - 1 = 3a\\$

$\sf:\implies \red {\:a \: = \: - \: \dfrac{1}{3} - - (3)}\\\\$

❍On SUBSTITUTING 'x = - 0' in equation (2), we get :-

$\sf :\implies\: 0 = a(0 - 0 + 1) + c(0 + 1)\\$

$\sf :\implies \:a + c = 0\\$

$\sf :\implies\:c = - \: a\\$

$\sf:\implies \red {\:\:c \: = \: \dfrac{1}{3} - - (4)}\\\\$

❍On substituting 'x = 1' in equation (2), we get :-

$\sf :\implies \:1 = a(1 - 1 + 1) + (b + c)(2)\\$

$\sf :\implies \:1 = a + 2b + 2c\\$

$\sf :\implies \:1 = - \: \dfrac{1}{3} + \dfrac{2}{3} + 2b\\$

$\sf :\implies \:1 = \dfrac{1}{3} + 2b\\$

$\sf :\implies \:2b = 1 - \dfrac{1}{3} \\$

$\sf :\implies \:2b = \dfrac{2}{3}\\$

$\sf:\implies \red {\:b \: = \: \dfrac{1}{3} \: - - (5)}\\$

❍Now, substitute the VALUES..

$\sf \: \: \dfrac{x}{(x + 1)( {x}^{2} - x + 1) } = \sf \: \dfrac{ - 1}{3(x + 1)} + \dfrac{x + 1}{3( {x}^{2} - x + 1) }\\$

$\sf \: \: \purple{\dfrac{x}{ {x}^{3} + 1 } } = \sf \pink {\: \dfrac{ - 1}{3(x + 1)} + \dfrac{x + 1}{3( {x}^{2} - x + 1) }}\\\\$

⠀⠀⠀ ━━━━━━━━━━━━━━━━━━━⠀

$\sf \green {Do \: you \: know \:?¿}$

Partial fraction decomposition is the breaking down of a rational expression into simpler parts. It is the opposite of adding rational expressions. When adding two rational expressions, there has to be a common denominator.

$\begin{gathered}\boxed{\begin{array}{c|c} \sf \pink { Term \: in \: denominator }& \sf\pink { Partial \: fraction \: decomposition }\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ax + b & \sf \displaystyle \sf \dfrac{A}{{ax + b}} \\ \\ \sf {(ax + b)}^{2} & \sf \displaystyle \sf \dfrac{{{A_1}}}{{ax + b}} + \sf\frac{{{A_2}}}{{{{\sf\left( {ax + b} \right)}^2}}} \\ \\ \sf {ax}^{2} + bx + c & \sf \displaystyle \sf \dfrac{{Ax + B}}{{a{x^2} + bx + c}} \end{array}} \\ \end{gathered}$

⠀⠀⠀ ━━━━━━━━━━━━━━━━━━━⠀

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