Question
Using binomial theorem write down the expansions of the following:
$\Big(\text{ax}-\frac{\text{b}}{\text{x}}\Big)^6$
Using binomial theorem write down the expansions of the following:
$\Big(\text{ax}-\frac{\text{b}}{\text{x}}\Big)^6$
The expansion of (x + y)n has n + 1 terms so the expansion of $\Big(\text{ax}-\frac{\text{b}}{\text{x}}\Big)^6$has 7 term Using binomial theorem to expand, we get
$\Big(\text{ax}-\frac{\text{b}}{\text{x}}\Big)^6={^6\text{C}}_0(\text{ax})^6\Big(\frac{\text{b}}{\text{x}}\Big)-{^6\text{C}}_1(\text{ax})^5\Big(\frac{\text{b}}{\text{x}}\Big)+{^6\text{C}}_2(\text{ax})^4\Big(\frac{\text{b}}{\text{x}}\Big)^2\\-{^6\text{C}}_3(\text{ax})^3\Big(\frac{\text{b}}{\text{x}}\Big)^3+{^6\text{C}}_4(\text{ax})^2\Big(\frac{\text{b}}{\text{x}}\Big)^4-{^6\text{C}}_5(\text{ax})\Big(\frac{\text{b}}{\text{x}}\Big)^5+{^6\text{C}}_6(\text{ax})^0\Big(\frac{\text{b}}{\text{x}}\Big)^6$
$=\text{a}^6\text{x}^6-6\text{a}^5\text{x}^5\frac{\text{b}}{\text{x}}+15\text{a}^4\text{x}^4\frac{\text{b}^2}{\text{x}^2}\\-20\text{a}^3\text{b}^3+15\text{a}^2\frac{\text{b}^4}{\text{x}^2}-6\text{a}\frac{\text{b}^5}{\text{x}}^4+\frac{\text{b}^6}{\text{x}^6}$
$=\text{a}^6\text{x}^6-6\text{a}^5\text{x}^4\text{b}\frac{\text{b}}{\text{x}}+15\text{a}^4\text{b}^2\text{x}^2-20\text{a}^3\text{b}^3+15\frac{\text{a}^2\text{b}^4}{\text{x}^2}-6\frac{\text{ab}^5}{\text{x}^4}+\frac{\text{b}^6}{\text{x}^6}$
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