Question

Resolve into partial fractions
$x \div {x}^{3} + 1$
​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Resolve \: in \: to \: partial \: fraction : \: \dfrac{x}{ {x}^{3} + 1}$

$\large\underline{\bold{Solution-}}$

We know,

$\sf \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )$

So,

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

Therefore,

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

Now,

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

On taking LCM, we get

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

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

$\rm :\longmapsto\: - 1 = a(1 + 1 + 1) + 0$

$\rm :\longmapsto\: - 1 = 3a$

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

On substituting 'x = - 0' in equation (2), we get

$\rm :\longmapsto\: 0 = a(0 - 0 + 1) + c(0 + 1)$

$\rm :\longmapsto\:a + c = 0$

$\rm :\longmapsto\:c = - \: a$

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

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

$\rm :\longmapsto\:1 = a(1 - 1 + 1) + (b + c)(2)$

$\rm :\longmapsto\:1 = a + 2b + 2c$

$\rm :\longmapsto\:1 = - \: \dfrac{1}{3} + \dfrac{2}{3} + 2b$

$\rm :\longmapsto\:1 = \dfrac{1}{3} + 2b$

$\rm :\longmapsto\:2b = 1 - \dfrac{1}{3}$

$\rm :\longmapsto\:2b = \dfrac{2}{3}$

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

Hence,

On substituting the values of a, b and c in equation (1), we get

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

Therefore,

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

Additional Information :-

Partial Fraction :-

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

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