Chapter 9.5, Problem 48E

To determine

The co-efficient $a$ of the term $a{x}^3$ in the Binomial expansion of ${\left(x-2\right)}^6$ .

Answer to Problem 48E

The co-efficient $a$ of the term $a{x}^3$ is $-160$ .

Explanation of Solution

Given information :

Given expression is ${\left(x-2\right)}^6$

We know that the ${\left(r+1\right)}^{th}$ term of the Binomial expansion of ${\left(x+y\right)}^n$ is ${}_n{C}_r{x}^{n-r}{y}^r$ ,

Thus, the ${\left(r+1\right)}^{th}$ term of the Binomial expansion of ${\left(x-2\right)}^6$ is ${}_6{C}_r{x}^{6-r}{\left(-2\right)}^r\left(\because n=6,x=x,y=-2\right)$

Now replace $r=3$ , to get $x^3$ in above expression

  $\begin{array}{l}{\Rightarrow }_6{C}_3{x}^{6-3}{\left(-2\right)}^3\\ \Rightarrow \frac{6!}{\left(6-3\right)!\left(3!\right)}{x}^3{\left(-2\right)}^3\\ \Rightarrow \frac{6\times 5\times 4\times 3!}{\left(3!\right)\times \left(3\times 2\times 1\right)}{x}^3{\left(-2\right)}^3\\ \Rightarrow 20\times \left(-8\right)\times x^3\\ \therefore -160x^3\end{array}$

Hence, the co-efficient $a$ of the term $a{x}^3$ is $-160$ .