Chapter 8.4, Problem 30E

To determine

To Simplify: To simplify and expand the given expression using binomial theorem.

Answer to Problem 30E

The answer is $x^6+6x^{5}y+15x^4{y}^2+20x^3{y}^3+15x^2{y}^4+6x{y}^5+y^6$

Explanation of Solution

Given information:Given expression ${(x+y)}^6$

Formula used: Binomial theorem ${(a+b)}^n=C{}_0{a}^n{b}^0+C{}_1{a}^{n-1}{b}^1+C{}_2{a}^{n-2}{b}^2+\mathrm{...}+C{}_n{a}^0{b}^n$

Calculation:

Since, the given expression is ${(x+y)}^6$

Using binomial theorem,

  $\begin{array}{l}{(x+y)}^6=C{}_0{x}^6{y}^0+C{}_1{x}^{6-1}{y}^1+C{}_2{x}^{6-2}{y}^2+C{}_3{x}^{6-3}{y}^3+C{}_4{x}^{6-4}{y}^4+C{}_5{x}^{6-5}{y}^5+C{}_6{x}^0{y}^6\\ {(x+y)}^6=\frac{6!}{0!\left(6-0\right)!}{x}^6+\frac{6!}{1!\left(6-1\right)!}{x}^{5}y+\frac{6!}{2!\left(6-2\right)!}{x}^4{y}^2+\frac{6!}{3!\left(6-3\right)!}{x}^3{y}^3\\ +\frac{6!}{4!\left(6-4\right)!}{x}^2{y}^4+\frac{6!}{5!\left(6-5\right)!}x{y}^5+\frac{6!}{6!\left(6-6\right)!}{y}^6\\ {(x+y)}^6=\frac{6!}{6!}{x}^6+\frac{6!}{1!5!}{x}^{5}y+\frac{6!}{2!4!}{x}^4{y}^2+\frac{6!}{3!3!}{x}^3{y}^3+\frac{6!}{4!2!}{x}^2{y}^4+\frac{6!}{5!1!}x{y}^5+\frac{6!}{6!}{y}^6\\ {(x+y)}^6=x^6+6x^{5}y+15x^4{y}^2+20x^3{y}^3+15x^2{y}^4+6x{y}^5+y^6\end{array}$

The results are verified from the graphing utility.

Hence the required expanded expression is $x^6+6x^{5}y+15x^4{y}^2+20x^3{y}^3+15x^2{y}^4+6x{y}^5+y^6$