Chapter 9, Problem 21RE

Solution

To identify: The expansion of the binomial ${\left(3x^2+y^3\right)}^5$ .

The expanded form of the binomial is

  ${\left(3x^2+y^3\right)}^5=243x^{10}+405x^8{y}^3+270x^6{y}^6+90x^4{y}^9+15x^2{y}^{12}+y^{15}$

Given information:

The given expression is ${\left(3x^2+y^3\right)}^5$ .

Explanation:

Consider the given expression

The binomial Theorem state that for any positive integer

  $n,$

  ${\left(a+b\right)}^n=\left(\begin{array}{l}n\\ 0\end{array}\right)a^n+\left(\begin{array}{l}n\\ 1\end{array}\right)a^{n-1}b+\mathrm{......}+\left(\begin{array}{l}n\\ r\end{array}\right)a^{n-r}{b}^r+\mathrm{.....}+\left(\begin{array}{l}n\\ n\end{array}\right)b^n$

Where $a=3x^2,n=5\text{and}b=y^3$

  $\begin{array}{c}{\left(3x^2+y^3\right)}^5=\left(\begin{array}{l}5\\ 0\end{array}\right){\left(3x^2\right)}^5+\left(\begin{array}{l}5\\ 1\end{array}\right){\left(3x^2\right)}^4\left(y^3\right)+\left(\begin{array}{l}5\\ 2\end{array}\right){\left(3x^2\right)}^3{\left(y^3\right)}^2+\left(\begin{array}{l}5\\ 3\end{array}\right){\left(3x^2\right)}^2{\left(y^3\right)}^3+\left(\begin{array}{l}5\\ 4\end{array}\right)\left(3x^2\right){\left(y^3\right)}^4\\ +\left(\begin{array}{l}5\\ 5\end{array}\right){\left(y^3\right)}^5\\ {\left(3x^2+y^3\right)}^5=1\left(243x^{10}\right)+5\left(81x^8\right)\left(y^3\right)+10\left(27x^6\right)\left(y^6\right)+10\left(9x^4\right)\left(y^9\right)+5\left(3x^2\right)\left(y^{12}\right)\\ +1\left(y^{15}\right)\\ {\left(3x^2+y^3\right)}^5=243x^{10}+405x^8{y}^3+270x^6{y}^6+90x^4{y}^9+15x^2{y}^{12}+y^{15}\end{array}$ .

So, the expression.

  ${\left(3x^2+y^3\right)}^5=243x^{10}+405x^8{y}^3+270x^6{y}^6+90x^4{y}^9+15x^2{y}^{12}+y^{15}$