Chapter 12, Problem 79RE

To determine

Expand the given expression ${\left(1-x^2\right)}^6$ .

Answer to Problem 79RE

  ${\left(1-x^2\right)}^6=1+6-x^2+15-x^4+20-x^6+15-x^8+6-x^{10}-x^{12}$

Explanation of Solution

Given information:

It is given in the question that, the expression ${\left(1-x^2\right)}^6$ .

Formula used:

  ${\left(a+b\right)}^6=a^6+6a^5{b}^1+15a^4{b}^2+20a^3{b}^3+15a^2{b}^4+6a{b}^5+b^6$

Calculation:

For finding expression ${\left(1-x^2\right)}^6$ ,

Put $a=1andb=-x^2$ in ${\left(a+b\right)}^6=a^6+6a^5{b}^1+15a^4{b}^2+20a^3{b}^3+15a^2{b}^4+6a{b}^5+b^6$

  $\begin{array}{l}{\left(1+\left(-x^2\right)\right)}^6=1^6+6*1^5{\left(-x^2\right)}^1+15*1^4{\left(-x^2\right)}^2+20*1^3{\left(-x^2\right)}^3+15*1^2{\left(-x^2\right)}^4+6*1*{\left(-x^2\right)}^5+{\left(-x^2\right)}^6\\ {\left(1-x^2\right)}^6=1+6-x^2+15-x^4+20-x^6+15-x^8+6-x^{10}-x^{12}\end{array}$