Chapter 4.2, Problem 49E

To determine

To find: the product of given expression using two different methods, and to find the method which is prefer.

Answer to Problem 49E

The product is $9a^8+66a^2{b}^2+97a^4{b}^4-88a^2{b}^6+16b^8$

Method 1 is preferable because it requires fewer steps than method 2.

Explanation of Solution

Given information:

Given expression

  ${\left(a^2+4b^2\right)}^2{\left(3a^2-b^2\right)}^2$

Calculation:

Consider the product ${\left(a^2+4b^2\right)}^2{\left(3a^2-b^2\right)}^2$

Method 1:

  $\begin{array}{c}{\left(a^2+4b^2\right)}^2{\left(3a^2-b^2\right)}^2=\left(a^4+8a^2{b}^2+16b^4\right)\left(9a^4-6a^2{b}^2+b^4\right)\\ =9a^4\left(a^4+8a^2{b}^2+16b^4\right)-6a^2{b}^2\left(a^4+8a^2{b}^2+16b^4\right)+b^2\left(a^4+8a^2{b}^2+16b^4\right)\\ =9a^8+72a^6{b}^2+144a^4{b}^4-6a^6{b}^2-48a^4{b}^4-96a^2{b}^6+a^4{b}^4+8a^2{b}^6+16b^8\\ =9a^8+66a^2{b}^2+97a^4{b}^4-88a^2{b}^6+16b^8\end{array}$

Therefore, the product is $9a^8+66a^2{b}^2+97a^4{b}^4-88a^2{b}^6+16b^8$ .

Method 2:

  $\begin{array}{c}{\left(a^2+4b^2\right)}^2{\left(3a^2-b^2\right)}^2={\left[\left(a^2+4b^2\right)\left(3a^2-b^2\right)\right]}^2\\ ={\left[\left(3a^4+11a^2{b}^2\right)-4b^4\right]}^2\\ ={\left(3a^4+11a^2{b}^2\right)}^2-8b^4\left(3a^4+11a^2{b}^2\right)+16b^8\\ =9a^8+121a^4{b}^4+66a^6{b}^2-24a^4{b}^4-88a^2{b}^6+16b^8\\ =9a^8+66a^2{b}^2+97a^4{b}^4-88a^2{b}^6+16b^8\end{array}$

Therefore, the product is $9a^8+66a^2{b}^2+97a^4{b}^4-88a^2{b}^6+16b^8$

Method 1 is preferable because it requires fewer steps than method 2.