Chapter 11.6, Problem 24PPS

To determine

The expansion of the binomial ${\left(y-3x\right)}^6$

Answer to Problem 24PPS

The fourth term is:

  $-540x^3{y}^3$

Explanation of Solution

Given information:

The given expression is as:

  ${\left(y-3x\right)}^6$

Concept used:

1. The binomial expansion for ${\left(x+y\right)}^n$ is as:

  ${\left(x+y\right)}^n={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{y}^r}$

Where the coefficient for term $x^{n-r}{y}^r$ is $\left(\begin{array}{c}n\\ r\end{array}\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

Since, the given expression is as:

  ${\left(y-3x\right)}^6$

On comparing this with ${\left(x+y\right)}^n$ , the value of $n$ is 6

  $\Rightarrow n=6$

On employing binomial expansion, we can express given expression as

  ${\left(y-3x\right)}^6={\displaystyle \sum _{r=0}^6\left(\begin{array}{c}6\\ r\end{array}\right)y^{6-r}{\left(3x\right)}^r}$

Therefore, the expand for ${\left(y-3x\right)}^6$ is

  $\begin{array}{c}{\left(y-3x\right)}^6={\displaystyle \sum _{r=0}^6\left(\begin{array}{c}6\\ r\end{array}\right)y^{6-r}{\left(-3x\right)}^r}\\ =\frac{6!}{\left(6-0\right)!0!}{\left(y\right)}^{6-0}\cdot {\left(-3x\right)}^0+\frac{6!}{\left(6-1\right)!1!}{\left(y\right)}^{6-1}{\left(-3x\right)}^1+\frac{6!}{\left(6-2\right)!2!}{\left(y\right)}^{6-2}\cdot {\left(-3x\right)}^2\\ +\frac{6!}{\left(6-3\right)!3!}{\left(y\right)}^{6-3}\cdot {\left(-3x\right)}^3+\frac{6!}{\left(6-4\right)!4!}{\left(y\right)}^{6-4}\cdot {\left(-3x\right)}^4+\frac{6!}{\left(6-5\right)!5!}{\left(y\right)}^{6-5}\cdot {\left(-3x\right)}^5\\ +\frac{6!}{\left(6-6\right)!6!}{\left(y\right)}^{6-6}\cdot {\left(-3x\right)}^6\\ =729x^6-1458x^{5}y+1215x^4{y}^2-540x^3{y}^3+135x^2{y}^4-18x{y}^5+y^6\\ y=3x\end{array}$

Thus, the fourth term of the given binomial is,

  $-540x^3{y}^3$