Chapter 11, Problem 13CR

To determine

To select: The correct option which shows solution of fourth term in expansion ${\left(3x-y^2\right)}^6$

(a). $-540x^3{y}^6$

(b). $540x^3{y}^6$

(c). $-540x^3{y}^3$

(d). $-27x^3{y}^6$

Answer to Problem 13CR

The correct option is (a).

The solution of fourth term in expansion ${\left(3x-y^2\right)}^6$ is $-540x^3{y}^6$ .

Explanation of Solution

Given information:

  ${\left(3x-y^2\right)}^6$

Calculation

Consider the equation:

  ${\left(3x-y^2\right)}^6$

Apply the binomial theorem:

  $a=3x,b=-y^2$

Use the formula:

  $\left(\frac{n}{i}\right)=\frac{n!}{i!\left(n-i\right)!}$

Put the values:

  $\begin{array}{l}i=0:\frac{6!}{0!\left(6-0\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^0\mathrm{...}\left(1\right)\\ i=1:\frac{6!}{1!\left(6-1\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^1\mathrm{...}\left(2\right)\\ i=2:\frac{6!}{2!\left(6-2\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^2\mathrm{...}\left(3\right)\\ i=3:\frac{6!}{3!\left(6-3\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^3\mathrm{...}\left(4\right)\end{array}$

  ${\left(3x-y^2\right)}^6=\left[\frac{6!}{0!\left(6-0\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^0\right]+\left[\frac{6!}{1!\left(6-1\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^1\right]+\left[\frac{6!}{2!\left(6-2\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^2\right]+\left[\frac{6!}{3!\left(6-3\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^3\right]$ Simplify equation (1):

  $\begin{array}{l}{\left(3x-y^2\right)}^6=\frac{6!}{0!\left(6-0\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^0\\ {\left(3x-y^2\right)}^6=729x^6\mathrm{...}\left(5\right)\end{array}$

Simplify equation (2):

  $\begin{array}{l}{\left(3x-y^2\right)}^6=\frac{6!}{1!\left(6-1\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^1\\ {\left(3x-y^2\right)}^6=-1458x^5{y}^2\mathrm{...}\left(6\right)\end{array}$

Simplify equation (3):

  $\begin{array}{l}{\left(3x-y^2\right)}^6=\frac{6!}{2!\left(6-2\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^2\\ {\left(3x-y^2\right)}^6=1215x^4{y}^4\mathrm{...}\left(7\right)\end{array}$

Simplify equation (4):

  $\begin{array}{l}{\left(3x-y^2\right)}^6=\frac{6!}{3!\left(6-3\right)!}{\left(3x\right)}^6{\left(-y^3\right)}^3\\ {\left(3x-y^2\right)}^6=-540x^3{y}^6\mathrm{...}\left(8\right)\end{array}$

The correct option is (a).

The solution of fourth term in expansion ${\left(3x-y^2\right)}^6$ is $-540x^3{y}^6$ .