Chapter 11.8, Problem 4ST

To determine

To calculate:

Simplify the specified term

Answer to Problem 4ST

The simplification of the fourth term is $-35a^8{b}^3$ .

Explanation of Solution

Given information:

Expression:

  ${\left(a^2-b\right)}^7$

Formula used:

Binomial theorem:

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

Where,

  $a=$ first term

  $b=$ second term

Calculation:

Consider expression ${\left(a^2-b\right)}^7$

Apply the binomial theorem for ${\left(a^2-b\right)}^7$

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

Where,

  $a=a^2,b=-b$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ ={\displaystyle {\sum }_{i=0}^7\left(\begin{array}{c}7\\ i\end{array}\right)}{\left(a^2\right)}^{\left(7-i\right)}{\left(-b\right)}^i\end{array}$

The expand summation are as follows:

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

The fourth term is,

  $\begin{array}{l}\frac{7!}{3!\left(7-3\right)!}{\left(a^2\right)}^4{\left(-b\right)}^3\\ \frac{7!}{3!\left(7-3\right)!}=\frac{210}{3!}\\ =\frac{210}{3!}{\left(a^2\right)}^4{\left(-b\right)}^3\\ {\left(a^2\right)}^4=a^8\\ =\frac{210}{3!}{a}^8{\left(-b\right)}^3\\ {\left(-b\right)}^3=-b^3\\ =\frac{210}{3!}{a}^8{\left(-b\right)}^3\end{array}$

Now remove paraeneses:

  $\begin{array}{l}\left(-a\right)=-a\\ =-\frac{210}{3!}{a}^8{b}^3\end{array}$

Multiply fractions:

  $\begin{array}{l}a.\frac{b}{c}=\frac{a\cdot b}{c}\\ =-\frac{210a^8{b}^3}{3!}=3!=6\\ =-\frac{210a^8{b}^3}{6}\\ =\frac{210}{6}\\ =35\\ =-35a^8{b}^3\end{array}$

The answer of the fourth term is $-35a^8{b}^3$ .