Question
Using binomial theorem, write down the expansions of the following:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^6$

Answer

The expansion of $(x + y)^n$ has $n + 1$ term so,
the expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^6$ has $7$ terms.
Using binomial theorem, we have
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^6={^6\text{C}}_0\text{x}^6\Big(\frac{1}{\text{x}}\Big)^0-{^6\text{C}}_1\text{x}^5\Big(\frac{1}{\text{x}}\Big)+{^6\text{C}}_2\text{x}^4\Big(\frac{1}{x}\Big)^2\\-{^6\text{C}_3\text{x}^3\Big(\frac{1}{\text{x}}\Big)}^3+{^6\text{C}_4\text{x}^2\Big(\frac{1}{\text{x}}\Big)}^4-{^6\text{C}_5\text{x}\Big(\frac{1}{\text{x}}\Big)}^5+{^6\text{C}_6\text{x}^0\Big(\frac{1}{\text{x}}\Big)}^6$
$\text{x}^6-6\text{x}^4+15\text{x}^2-20+\frac{15}{\text{x}^2}-\frac{6}{\text{x}^4}+\frac{1}{\text{x}^6}$

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