Chapter 2.3, Problem 19E

To determine

Expand ${\left(a_1^2+2a_2+a_3\right)}^3$ using multinomial theorem.

Explanation of Solution

Calculation:

Multinomial theorem:

The multinomial theorem for all the non-negative integers $n_1,n_2,\mathrm{...}{n}_k$ is,

${\left(a_1+\mathrm{...}+a_r\right)}^n={\displaystyle \sum _{n_1+\mathrm{...}+n_r=n}\left(\begin{array}{c}n\\ n_1,n_2,\mathrm{...},n_r\end{array}\right)a_1^{n_1}{a}_2^{n_2}\mathrm{...}{a}_r^{n_r}}$

In the formula, $\left(\begin{array}{c}n\\ n_1,n_2,\mathrm{...},n_r\end{array}\right)=\frac{n!}{n_1!n_2!\mathrm{...}{n}_r!}$

Consider,

${\left(a_1^2+2a_2+a_3\right)}^3=\left[\begin{array}{l}\left(\begin{array}{c}3\\ 0,0,3\end{array}\right){\left(a_1^2\right)}^0{\left(2a_2\right)}^0\left(a_3^3\right)+\left(\begin{array}{c}3\\ 0,1,2\end{array}\right){\left(a_1^2\right)}^0{\left(2a_2\right)}^1\left(a_3^2\right)+\\ \left(\begin{array}{c}3\\ 0,2,1\end{array}\right){\left(a_1^2\right)}^0{\left(2a_2\right)}^2\left(a_3^1\right)+\left(\begin{array}{c}3\\ 0,3,0\end{array}\right){\left(a_1^2\right)}^0{\left(2a_2\right)}^3\left(a_3^0\right)+\\ \left(\begin{array}{c}3\\ 1,0,2\end{array}\right){\left(a_1^2\right)}^1{\left(2a_2\right)}^0\left(a_3^2\right)+\left(\begin{array}{c}3\\ 1,1,1\end{array}\right){\left(a_1^2\right)}^1{\left(2a_2\right)}^1\left(a_3^1\right)+\\ \left(\begin{array}{c}3\\ 1,2,0\end{array}\right){\left(a_1^2\right)}^1{\left(2a_2\right)}^2\left(a_3^0\right)+\left(\begin{array}{c}3\\ 2,0,1\end{array}\right){\left(a_1^2\right)}^2{\left(2a_2\right)}^0\left(a_3^1\right)+\\ \left(\begin{array}{c}3\\ 2,1,0\end{array}\right){\left(a_1^2\right)}^2{\left(2a_2\right)}^1\left(a_3^0\right)+\left(\begin{array}{c}3\\ 3,0,0\end{array}\right){\left(a_1^2\right)}^3{\left(2a_2\right)}^0\left(a_3^0\right)\end{array}\right]$

$=\left[\begin{array}{l}\frac{3!}{0!0!3!}\left(1\right)\left(1\right)\left(a_3^3\right)+\frac{3!}{0!1!2!}\left(1\right)\left(2a_2\right)\left(a_3^2\right)+\\ \frac{3!}{0!2!1!}\left(1\right)\left(4a_2^2\right)\left(a_3^1\right)+\frac{3!}{0!3!0!}\left(1\right)\left(8a_2^3\right)\left(1\right)+\\ \frac{3!}{1!0!2!}\left(a_1^2\right)\left(1\right)\left(a_3^2\right)+\frac{3!}{1!1!1!}\left(a_1^2\right)\left(2a_2\right)\left(a_3^1\right)+\\ \frac{3!}{1!2!0!}\left(a_1^2\right)\left(4a_2^2\right)\left(1\right)+\frac{3!}{2!0!1!}\left(a_1^4\right)\left(1\right)\left(a_3^1\right)+\\ \frac{3!}{2!1!0!}\left(a_1^4\right)\left(2a_2\right)\left(1\right)+\frac{3!}{3!0!0!}\left(a_1^6\right)\left(1\right)\left(1\right)\end{array}\right]$

$\begin{array}{l}=\left[\begin{array}{l}{a}_3^3+3\left(2a_2\right)\left(a_3^2\right)+3\left(4a_2^2\right)\left(a_3\right)+\left(8a_2^3\right)+\\ 3\left(a_1^2\right)\left(a_3^2\right)+6\left(a_1^2\right)\left(2a_2\right)\left(a_3\right)+3\left(a_1^2\right)\left(4a_2^2\right)\\ +3\left(a_1^4\right)\left(a_3^1\right)+3\left(a_1^4\right)\left(2a_2\right)+\left(a_1^6\right)\end{array}\right]\\ =a_3^3+6a_2{a}_3^2+12a_2^2{a}_3+8a_2^3+3a_1^2{a}_3^2+12a_1^2{a}_2{a}_3+12a_1^2{a}_2^2+3a_1^4{a}_3^1+6a_1^4{a}_2+a_1^6\end{array}$

Hence, the term ${\left(a_1^2+2a_2+a_3\right)}^3$ is expanded as,

$a_3^3+6a_2{a}_3^2+12a_2^2{a}_3+8a_2^3+3a_1^2{a}_3^2+12a_1^2{a}_2{a}_3+12a_1^2{a}_2^2+3a_1^4{a}_3^1+6a_1^4{a}_2+a_1^6$