Chapter 4.2, Problem 1E

To determine

To describe: the different methods to expand ${\left(x+3\right)}^3$ .

Answer to Problem 1E

Method 1:

Pascal triangle

Method 2:

Formula of ${\left(a+b\right)}^3=a^3+b^3+3a^{2}b+3a{b}^2$

Method 3:

Horizontal product

Explanation of Solution

Given information:

An expression is given as

  ${\left(x+3\right)}^3$

Calculation:

Three method to expand ${\left(x+3\right)}^3$

is described below:

Method 1:

Pascal triangle:

Here, $n=3$ .

First write out the Pascal’s triangle such that the row that begins with 1, 3.

  $\begin{array}{ccccccc}& & & 1& & & \\ & & 1& & 1& & \\ & 1& & 2& & 1& \\ 1& & 3& & 3& & 1\end{array}$

From the Pascal’s triangle, it is observed that the binomial coefficients of the expression are 1, 3, 3 and 1.

Expand the expression ${\left(x+3\right)}^3$ by using formula in Binomial theorem as follows.

  $\begin{array}{c}{\left(x+3\right)}^3=\left(1\right){\left(x\right)}^3+3{\left(x\right)}^{3-1}\left(3\right)+3{\left(x\right)}^{3-2}{\left(3\right)}^2+1{\left(3\right)}^3\\ =x^3+9x^2+27x+27\end{array}$

Method 2:

Formula of ${\left(a+b\right)}^3=a^3+b^3+3a^{2}b+3a{b}^2$ :

  $\begin{array}{c}{\left(x+3\right)}^3={\left(x\right)}^3+3{\left(x\right)}^{2}3+3x{\left(3\right)}^2+3^3\\ =x^3+9x^2+27x+27\end{array}$

Method 3:

Horizontal product:

  $\begin{array}{c}{\left(x+3\right)}^3=\left(x+3\right)\left(x+3\right)\left(x+3\right)\\ =\left(x^2+3x+3x+9\right)\left(x+3\right)\\ =\left(x^2+6x+9\right)\left(x+3\right)\\ =x^3+3x^2+6x^2+18x+9x+27\\ =x^3+9x^2+27x+27\end{array}$