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Question
Using binomial theorem, write down the expansions :
(ix) \[\left( x + 1 - \frac{1}{x} \right)\]
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Solution
(ix) \[(x + 1 - \frac{1}{x} )^3 \]
\[ = ^{3}{}{C}_0 (x + 1 )^3 (\frac{1}{x} )^0 - ^{3}{}{C}_1 (x + 1 )^2 (\frac{1}{x} )^1 + ^{3}{}{C}_2 (x + 1 )^1 (\frac{1}{x} )^2 - ^{3}{}{C}_3 (x + 1 )^0 (\frac{1}{x} )^3\]
\[= (x + 1 )^3 - 3(x + 1 )^2 \times \frac{1}{x} + 3\frac{x + 1}{x^2} - \frac{1}{x^3}\]
\[ = x^3 + 1 + 3x + 3 x^2 - \frac{3 x^2 + 3 + 6x}{x} + 3\frac{x + 1}{x^2} - \frac{1}{x^3}\]
\[ = x^3 + 1 + 3x + 3 x^2 - 3x - \frac{3}{x} - 6 + \frac{3}{x} + \frac{3}{x^2} - \frac{1}{x^3}\]
\[ = x^3 + 3 x^2 - 5 + \frac{3}{x^2} - \frac{1}{x^3}\]
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