Chapter 12.5, Problem 23AYU

In Problems 17-28, expand each expression using the Binomial Theorem.

${\left(x^2+y^2\right)}^5$

To determine

To find: The expansion of the given binomial expression ${\left(x^2 + y^2\right)}^5$ using Binomial Theorem.

Answer to Problem 23AYU

Solution:

The expansion of the expression is $x^{10} + 5y^2{x}^8 + 10y^4{x}^6 + 10y^6{x}^4 + 5y^8{x}^2 + y^{10}$.

Explanation of Solution

Given:

Expression is given as ${\left(x^2 + y^2\right)}^5$.

Formula used:

The Binomial Theorem:

Let $x$ and $a$ be real numbers. For any positive integer $n$, we have

$\begin{array}{l}{\left(x + a\right)}^n = \left(\begin{array}{c}n\\ 0\end{array}\right)x^n + \left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n - 1} + \cdots + \left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n - j} + \cdots + \left(\begin{array}{c}n\\ n\end{array}\right)a^n\\ {\left(x + a\right)}^n = {\displaystyle \sum _{k = 0}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^{n - j}{a}^j}\end{array}$

Here $x = x^2$, $a = y^2$ and $n = 5$. Applying Binomial theorem,

$\begin{array}{l}{\left(x^2 + y^2\right)}^5 = \left(\begin{array}{c}5\\ 0\end{array}\right){\left(x^2\right)}^5 + \left(\begin{array}{c}5\\ 1\end{array}\right)\left(y^2\right){\left(x^2\right)}^4 + \left(\begin{array}{c}5\\ 2\end{array}\right){\left(y^2\right)}^2{\left(x^2\right)}^3 + \left(\begin{array}{c}5\\ 3\end{array}\right){\left(y^2\right)}^3{\left(x^2\right)}^2\\ + \left(\begin{array}{c}5\\ 4\end{array}\right){\left(y^2\right)}^4{x}^2 + \left(\begin{array}{c}5\\ 5\end{array}\right){\left(y^2\right)}^5\end{array}$

$\begin{array}{l} = 1.x^{10} + 5.\left(y^2\right)x^8 + 10.\left(y^4\right)x^6 + 10.\left(y^6\right)x^4 + 5.\left(y^8\right)x^2 + 1.y^{10}\\ {\left(x^2 + y^2\right)}^5 = x^{10} + 5y^2{x}^8 + 10y^4{x}^6 + 10y^6{x}^4 + 5y^8{x}^2 + y^{10}\end{array}$