Chapter 4.4, Problem 33WE

a)

To determine

To simplify the Sum of given monomials.

Answer to Problem 33WE

The sum of the given monomials is $17x^6.$

Explanation of Solution

Given:

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

Calculations: Putting plus sign in between given monomials

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

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

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

  $3^2{\left(x^3\right)}^2$

  $\Rightarrow 3^2{x}^6$

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

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

  $2^3{\left(x^2\right)}^3$

  $\Rightarrow 2^3{x}^6$

Therefore,

  ${\left(3x^3\right)}^2+{\left(2x^2\right)}^3$ = $3^2{x}^6+2^3{x}^6$

  $\Rightarrow 9x^6+8x^6$

Add like terms:

  $\Rightarrow 17x^6$

Therefore, the sum of the given monomials is $17x^6.$

(b)

To determine

The product of given monomials

Answer to Problem 33WE

The product of the monomials is $72x^{12}.$

Explanation of Solution

Given:

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

Calculations: Putting product sign in between given monomials

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

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

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

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

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

  $3^2{\left(x^3\right)}^2=3^2{x}^6$

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

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

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

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

  $2^3{\left(x^2\right)}^3=2^3{x}^6$

Therefore,

  ${\left(3x^3\right)}^2\cdot {\left(2x^2\right)}^3=3^2{x}^6\cdot 2^3{x}^6$

Apply exponent rule: $a^b\cdot a^c=a^{b+c}$

  $3^2{x}^6\cdot 2^3{x}^6=3^2\cdot 2^3{x}^{6+6}$

  $\Rightarrow 72x^{12}$

Therefore, the product of the monomials is $72x^{12}.$