Chapter 11.7, Problem 17WE

To determine

To calculate:

Expansion and simplification of expression

  ${\left(a+b\right)}^7+{\left(a-b\right)}^7$

Answer to Problem 17WE

The simplified expression is $2a^7+42a^5{b}^2+70a^3{b}^4+14a{b}^6$ .

Explanation of Solution

Given information:

The given expression is,

  ${\left(a+b\right)}^7+{\left(a-b\right)}^7$

Formula:

The formula of binomial theorem is used,

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\right)}{a}^{\left(n-1\right)}{b}^i$

Calculation:

The given expression is,

  ${\left(a+b\right)}^7+{\left(a-b\right)}^7$

Here, consider

  $\begin{array}{l}a=a\\ b=b\end{array}$

By using the binomial theorem of expansion. Expanding the above expression.

  $\begin{array}{l}=\frac{7!}{0!\left(7-0\right)!}{a}^7{b}^0+\frac{7!}{1!\left(7-1\right)!}{a}^6{b}^1+\frac{7!}{2!\left(7-2\right)!}{a}^5{b}^2+\frac{7!}{3!\left(7-3\right)!}{a}^4{b}^3+\\ \frac{7!}{4!\left(7-4\right)!}{a}^3{b}^4+\frac{7!}{1!\left(7-5\right)!}{a}^2{b}^5+\frac{7!}{1!\left(7-6\right)!}{a}^1{b}^6+\frac{7!}{1!\left(7-7\right)!}{a}^0{b}^7\\ \end{array}$

By simplifying each factor the expression obtained is,

  $=2a^7+42a^5{b}^2+70a^3{b}^4+14a{b}^6$