Question

X^2-6x/(x-3)^3 resolve into partial fraction

Answer 1

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

Given function is

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 6x}{ {(x - 3)}^{3} }$

To resolve this in to partial fraction, Let assume that

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 6x}{ {(x - 3)}^{3} } = \dfrac{a}{(x - 3)} + \dfrac{b}{ {(x - 3)}^{2} } + \dfrac{c}{ {(x - 3)}^{3} }$

On taking LCM, we get

$\rm :\longmapsto\: {x}^{2} - 6x = a {(x - 3)}^{2} + b(x - 3) + c - - - (1)$

On substituting x = 3 in equation (1), we get

$\rm :\longmapsto\: {3}^{2} - 18 = c$

$\rm :\longmapsto\: 9 - 18 = c$

$\bf\implies \:c = - 9$

On substituting x = 0 in equation (1), we get

$\rm :\longmapsto\: {0}^{2} -0 = a {(0 - 3)}^{2} + b(0 - 3) + c$

$\rm :\longmapsto\:0 = 9a - 3b + c$

$\rm :\longmapsto\:0 = 9a - 3b - 9$

$\rm :\longmapsto\:0 = 3a - b - 3$

$\bf\implies \:3a - b = 3 - - - - (2)$

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

$\rm :\longmapsto\: {1}^{2} - 6= a {(1 - 3)}^{2} + b(1 - 3) + c$

$\rm :\longmapsto\: - 5= 4a - 2b - 9$

$\rm :\longmapsto\:9 - 5= 4a - 2b$

$\rm :\longmapsto\:4a - 2b = 4$

$\bf\implies \:a - b = 2 - - - (3)$

On Subtracting equation (3) from equation (2), we get

$\rm :\longmapsto\:2a = 1$

$\bf\implies \:a = \dfrac{1}{2}$

On substituting the value of a, in equation (3), we get

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

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

$\rm :\longmapsto\:b = \dfrac{1 - 4}{2}$

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

So, given expression can be resolved as

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 6x}{ {(x - 3)}^{3} } = \dfrac{1}{2(x - 3)} - \dfrac{3}{ {2(x - 3)}^{2} } - \dfrac{9}{ {(x - 3)}^{3} }$