Question

Expand ${(\frac{x}{3}+\frac{1}{x})}^5$

Answer 1

Using Binomial theorem,
${(\frac{x}{3}+\frac{1}{x})}^5$
${=}^5{C}_0{\left(\frac{x}{3}\right)}^5{+}^5{C}_1{\left(\frac{x}{3}\right)}^4\left(\frac{1}{x}\right){+}^5{C}_2{\left(\frac{x}{3}\right)}^3{\left(\frac{1}{x}\right)}^2$

${+}^5{C}_3{\left(\frac{x}{3}\right)}^2{\left(\frac{1}{x}\right)}^3{+}^5{C}_4\left(\frac{x}{3}\right){\left(\frac{1}{x}\right)}^4{+}^5{C}_5{\left(\frac{1}{x}\right)}^5$
$=\frac{x^5}{243}+5\left(\frac{x^4}{81}\right)\left(\frac{1}{x}\right)+10\left(\frac{x^3}{27}\right)\left(\frac{1}{x^2}\right)+10\left(\frac{x^2}{9}\right)\left(\frac{1}{x^3}\right)+5\left(\frac{x}{3}\right)\left(\frac{1}{x^4}\right)+\frac{1}{x^5}$
$=\frac{x^3}{243}+\frac{5x^3}{81}+\frac{10x}{27}+\frac{10}{9x}+\frac{5}{3x^3}+\frac{1}{x^5}$