${{x + 1} \over {(x - 1)\,(x - 2)\,(x - 3)}} = $
IIT 1996, Easy
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(c) \({{x + 1} \over {(x - 1)\,(x - 2)\,(x - 3)}} = {A \over {x - 1}} + {B \over {x - 2}} + {C \over {x - 3}}\)
(c) \({{x + 1} \over {(x - 1)\,(x - 2)\,(x - 3)}} = {A \over {x - 1}} + {B \over {x - 2}} + {C \over {x - 3}}\)
\( \Rightarrow \)\(x + 1 = A\,(x - 2)\,(x - 3)\, + B(x - 1)\,(x - 3) + C(x - 1)\,(x - 2)\)
Putting \(x = 1,\,A = 1\) ; \(x=2\) gives \(B = - 3\),
For \(x = 3,\,C = 2\)
\(\therefore \) Given expression = \({1 \over {x - 1}} - {3 \over {x - 2}} + {2 \over {x - 3}}\).
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