7.1 Q-23

Exercise 7.1 — Question 23

Problem

Expand using the Binomial Theorem and simplify: ${\left(x-\frac{1}{x}\right)}^6$

Solution

Using the Binomial Theorem: $(a+b{)}^n={\sum }_{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})a^{n-r}{b}^r$

Here $a=x$, $b=-\frac{1}{x}$, and $n=6$.

${\left(x-\frac{1}{x}\right)}^6={\sum }_{r=0}^6\left(\genfrac{}{}{0ex}{}{6}{r}\right)x^{6-r}{\left(-\frac{1}{x}\right)}^r$

Each term is: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{6}{r}\right)x^{6-r}\cdot (-1{)}^r\cdot x^{-r}=(-1{)}^r\left(\genfrac{}{}{0ex}{}{6}{r}\right)x^{6-2r}$

Expanding term by term:

Final Answer

${\left(x-\frac{1}{x}\right)}^6=x^6-6x^4+15x^2-20+\frac{15}{x^2}-\frac{6}{x^4}+\frac{1}{x^6}$

Key Steps

1. Identify $a$, $b$, and $n$ from the expression.
2. Apply the general term formula $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$.
3. Simplify powers of $x$ by combining $x^{6-r}\cdot x^{-r}=x^{6-2r}$.
4. Account for the sign $(-1{)}^r$ from the negative term.