Chapter 4.4, Problem 35WE

a)

To determine

To simplify the Sum of given monomials.

Answer to Problem 35WE

The sum of the given monomials is $2a^4{b}^6.$

Explanation of Solution

Given:

We have given two monomials: $a^2{\left(a{b}^3\right)}^2,b^2{\left(a^2{b}^2\right)}^2$

Calculations: Put plus sign in between given monomials

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

Simplify: $a^2{\left(a{b}^3\right)}^2$

Apply exponent rule: $a^m{b}^m={\left(ab\right)}^m$

  $a^2{\left(a{b}^3\right)}^2={\left(aa{b}^3\right)}^2$

  $\Rightarrow {\left(a^2{b}^3\right)}^2$

Apply exponent rule: ${\left(a\cdot b\right)}^n=a^n{b}^n$

  ${\left(a^2{b}^3\right)}^2={\left(a^2\right)}^2{\left(b^3\right)}^2$

  $\Rightarrow a^4{b}^6$

Simplify: $b^2{\left(a^2{b}^2\right)}^2$

Apply exponent rule: $a^m{b}^m={\left(ab\right)}^m$

  $b^2{\left(a^2{b}^2\right)}^2={\left(a^2{b}^{2}b\right)}^2$

  $\Rightarrow {\left(a^2{b}^3\right)}^2$

Apply exponent rule: ${\left(a\cdot b\right)}^n=a^n{b}^n$

  ${\left(a^2{b}^3\right)}^2={\left(a^2\right)}^2{\left(b^3\right)}^2$

  $\Rightarrow a^4{b}^6$

Therefore,

  $a^2{\left(a{b}^3\right)}^2+b^2{\left(a^2{b}^2\right)}^2=a^4{b}^6+a^4{b}^6$

Add like terms:

  $a^4{b}^6+a^4{b}^6=2a^4{b}^6$

Therefore, the sum of the given monomials is $2a^4{b}^6.$

(b)

To determine

The product of given monomials

Answer to Problem 35WE

The product of the monomials is $a^8{b}^{12}.$

Explanation of Solution

Given:

We have given two monomials: $a^2{\left(a{b}^3\right)}^2,b^2{\left(a^2{b}^2\right)}^2$

Calculation: Put product sign in between given monomials

  $a^2{\left(a{b}^3\right)}^2\cdot b^2{\left(a^2{b}^2\right)}^2$

Apply exponent rule: $a^m{b}^m={\left(ab\right)}^m$

  $a^2{\left(a{b}^3\right)}^2\cdot b^2{\left(a^2{b}^2\right)}^2={\left(a^{2}aa{b}^3{b}^{2}b\right)}^2$

  $\Rightarrow {\left(a^4{b}^6\right)}^2$

Apply exponent rule: ${\left(a\cdot b\right)}^n=a^n{b}^n$

  ${\left(a^4{b}^6\right)}^2={\left(a^4\right)}^2{\left(b^6\right)}^2$

  $=a^8{\left(b^6\right)}^2$

  $=a^8{b}^{12}$

Therefore, the product of the monomials is $a^8{b}^{12}.$