[tex]\large\underline{\sf{Solution-}}[/tex]
Given expression is
[tex]\sf \: \dfrac{1}{(x - 1)( {x}^{2} - 9)} \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: \dfrac{1}{(x - 1)( {x}^{2} - {3}^{2} )} \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{(x - 1)(x - 3)(x + 3)} \\ \\ [/tex]
Let assume that
[tex]\sf \: \dfrac{1}{(x - 1)(x - 3)(x + 3)} = \dfrac{a}{x - 1} + \dfrac{b}{x - 3} + \dfrac{c}{x + 3} - - - (1) \\ \\ [/tex]
On taking LCM, we get
[tex]\sf \: 1 = a(x - 3)(x + 3) + b(x - 1)(x + 3) + c(x - 1)(x - 3) \\ \\ [/tex]
On substituting x = 1, we get
[tex]\sf \: 1 = a(1 - 3)(1 + 3) + b(1 - 1)(1 + 3) + c(1 - 1)(1 - 3) \\ \\ [/tex]
[tex]\sf \: 1 = a( - 2)(4) +0 + 0 \\ \\ [/tex]
[tex]\bf\implies \:a \: = \: - \: \dfrac{1}{8} \\ \\ [/tex]
On substituting x = 3, we get
[tex]\sf \: 1 = a(3 - 3)(3 + 3) + b(3 - 1)(3 + 3) + c(3 - 1)(3 - 3) \\ \\ [/tex]
[tex]\sf \: 1 = 0 + b(2)(6)+ 0 \\ \\ [/tex]
[tex]\bf\implies \:b \: = \: \dfrac{1}{12} \\ \\ [/tex]
On substituting x = - 3, we get
[tex]\sf \: 1 = a( - 3 - 3)( - 3 + 3) + b( - 3 - 1)( - 3 + 3) + c( - 3 - 1)( - 3 - 3) \\ \\ [/tex]
[tex]\sf \: 1 = 0 + 0 + c( - 4)( - 6) \\ \\ [/tex]
[tex]\bf\implies \:c \: = \: \dfrac{1}{24} \\ \\ [/tex]
On substituting the values of a, b and c in equation (1), we get
[tex]\bf \: \dfrac{1}{(x - 1)(x - 3)(x + 3)} = \dfrac{ - 1}{8(x - 1)} + \dfrac{1}{12(x - 3)} + \dfrac{1}{24(x + 3)} \\ \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf Fraction & \bf Partial \: Fraction \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \dfrac{1}{(ax + b)(cx + d)} & \sf \dfrac{p}{ax + b} + \dfrac{q}{cx + d} \\ \\ \sf \dfrac{1}{ {(ax + b)}^{2} } & \sf \dfrac{p}{ax + b} + \dfrac{q}{ {(ax + b)}^{2} } \\ \\ \sf \dfrac{1}{(ax + b)(c {x}^{2} + d)} & \sf \dfrac{p}{ax + b} + \dfrac{qx + r}{c {x}^{2} + d} \end{array}} \\ \end{gathered} \\ [/tex]
