Chapter 8.4, Problem 41E

To determine

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

Answer to Problem 41E

The answer is $x^8+4x^6{y}^2+6x^4{y}^4+4x^2{y}^6+y^8$

Explanation of Solution

Given information:Given expression ${(x^2+y^2)}^4$

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 ${(x^2+y^2)}^4$

Using binomial theorem,

  $\begin{array}{}$ {(x^2+y^2)}^4=C{}_0{(x^2)}^4{(y^2)}^0+C{}_1{(x^2)}^{4-1}{(y^2)}^1+C{}_2{(x^2)}^{4-2}{(y^2)}^2+C{}_3{(x^2)}^{4-3}{(y^2)}^3+C{}_4{(x^2)}^0{(y^2)}^4$$ {(x^2+y^2)}^4=\frac{4!}{0!\left(4-0\right)!}{x}^8+\frac{4!}{1!\left(4-1\right)!}{x}^6{y}^2+\frac{4!}{2!\left(4-2\right)!}{x}^4{y}^4+\frac{4!}{3!\left(4-3\right)!}{x}^2{y}^6+\frac{4!}{4!\left(4-4\right)!}{y}^8$$ {(x^2+y^2)}^4=\frac{4!}{0!4!}{x}^8+\frac{4!}{1!3!}{x}^6{y}^2+\frac{4!}{2!2!}{x}^4{y}^4+\frac{4!}{3!1!}{x}^2{y}^6+\frac{4!}{4!0!}{y}^8$$ {(x^2+y^2)}^4=x^8+4x^6{y}^2+6x^4{y}^4+4x^2{y}^6+y^8$\end{array}$

The results are verified from the graphing utility.

Hence the required expanded expression is $x^8+4x^6{y}^2+6x^4{y}^4+4x^2{y}^6+y^8$