Chapter 11.7, Problem 4WE

To determine

To calculate:

The expansion and simplification of the expression.

  ${\left(p-q\right)}^6$

Answer to Problem 4WE

The simplified expression $p^6-6p^{5}q+15p^4{q}^2-20p^3{q}^3+15p^2{q}^4-6p{q}^5+q^6$ .

Explanation of Solution

Given information:

The given expression is,

  ${\left(p-q\right)}^6$

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,

  ${\left(p-q\right)}^6$

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 $\begin{array}{l}a=p\\ b=-q\end{array}$

  $={\displaystyle {\sum }_i^6=0\left({}_i^6\right)}{p}^{\left(6-i\right)}{\left(-q\right)}^i$

Now expand the above expression by binomial theorem and substituting the values in formula,

Now, the expression is written as,

  $=\frac{6!}{0!\left(6-0\right)}{p}^6{\left(-q\right)}^0+\frac{5!}{1!\left(6-0\right)}{p}^5{\left(-q\right)}^1+\frac{5!}{2!\left(6-0\right)}{p}^4{\left(-q\right)}^2+\frac{5!}{3!\left(6-0\right)}{p}^3{\left(-d\right)}^3$$+\frac{5!}{4!\left(6-0\right)}{p}^2{\left(-d\right)}^4+\frac{5!}{5!\left(6-0\right)}{p}^1{\left(-d\right)}^5+\frac{6!}{6!\left(6-0\right)}{p}^0{\left(-q\right)}^6$

By the simplification of the above expression the expression obtained is,

  $=\frac{6!}{0!\left(6-0\right)}{p}^6{\left(-q\right)}^0+\frac{5!}{1!\left(6-0\right)}{p}^5{\left(-q\right)}^1+\frac{5!}{2!\left(6-0\right)}{p}^4{\left(-q\right)}^2+\frac{5!}{3!\left(6-0\right)}{p}^3{\left(-d\right)}^3$$+\frac{5!}{4!\left(6-0\right)}{p}^2{\left(-d\right)}^4+\frac{5!}{5!\left(6-0\right)}{p}^1{\left(-d\right)}^5+\frac{6!}{6!\left(6-0\right)}{p}^0{\left(-q\right)}^6$

  $=p^6-6p^{5}q+15p^4{q}^2-20p^3{q}^3+15p^2{q}^4-6p{q}^5+q^6$