Chapter 11.7, Problem 18WE

To determine

To calculate:

The expansion and simplification of the expression.

  ${\left(a+b\right)}^8-{\left(a-b\right)}^8$

Answer to Problem 18WE

The simplified expression is $16a^7+112a^{5}b+112a^{3}b+16a{b}^7$

Explanation of Solution

Given information:

The given expression is,

  ${\left(a+b\right)}^8-{\left(a-b\right)}^8$

Formula:

The formula of binomial expression is used,

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

Calculation:

The expression given is,

. ${\left(a+b\right)}^8-{\left(a-b\right)}^8$

Now here by using the formula of binomial expansion,

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

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

  $={\displaystyle {\sum }_i^8=0\left({}_i^8\right)}\times {\left(a\right)}^{\left(8-i\right)}{}_{b^i}$

Now expand the above expression by binomial theorem.

  $=\frac{8!}{0!\left(8-0\right)!}\times {\left(a\right)}^8{\left(b\right)}^0+\frac{8!}{1!\left(8-0\right)!}\times {\left(a\right)}^7{\left(b\right)}^1+\frac{8!}{2!\left(8-0\right)!}\times {\left(a\right)}^6{\left(b\right)}^2+$

  $\frac{8!}{3!\left(8-0\right)!}\times {\left(a\right)}^5{\left(b\right)}^3+\frac{8!}{4!\left(8-0\right)!}\times {\left(a\right)}^6{\left(b\right)}^2+\frac{8!}{5!\left(8-0\right)!}\times {\left(a\right)}^7{\left(b\right)}^1+\frac{8!}{6!\left(8-0\right)!}\times {\left(a\right)}^8{\left(b\right)}^0$

By expanding the above brackets the simplified equation obtained is,

  $\begin{array}{l}=a^8+8a^{7}b+28a^6{b}^2+56a^5{b}^3+70a^4{b}^4+56a^3{b}^5+28a^2{b}^6+8a{b}^7+b^8-\\ \left(a^8-8a^{7}b+28a^6{b}^2-56a^5{b}^3+70a^4{b}^4-56a^3{b}^5+28a^2{b}^6-8a{b}^7+b^8\right)\\ =16a^7+112a^{5}b+112a^{3}b+16a{b}^7\end{array}$