Chapter 6.5, Problem 7CYU

To determine

To find: the factor of the given polynomial and check whether the polynomial is factored or prime.

Answer to Problem 7CYU

  $a^6{x}^2-b^6{x}^2=x^2\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)$ .

Explanation of Solution

Given:

The given polynomial is $a^6{x}^2-b^6{x}^2$ .

Calculation:

The given polynomial is $a^6{x}^2-b^6{x}^2$ .

Now, to find the factor of the given polynomial. Therefore,

  $\begin{array}{l}{a}^6{x}^2-b^6{x}^2\left\{\text{given polynomial}\right\}\\ =x^2\left(a^6-b^6\right)\left\{\text{factor out the GCF}\right\}\\ =x^2\left({\left(a^3\right)}^2-{\left(b^3\right)}^2\right)\\ \{a^2-b^2=\left(a+b\right)\left(a-b\right)\}\\ =x^2\left(a^3+b^3\right)\left(a^3-b^3\right)\end{array}$

Since are both perfect cubes, so write the factor of the sum of two cubes.

  $\begin{array}{l}{x}^2\left(a^6-b^6\right)\\ =x^2\left(a^3+b^3\right)\left(a^3-b^3\right)\\ \left\{\begin{array}{l}{a}^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\\ a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\end{array}\right\}\\ =x^2\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\end{array}$

Thus, $a^6{x}^2-b^6{x}^2=x^2\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)$

Hence, the factor of the given polynomial $a^6{x}^2-b^6{x}^2=x^2\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)$ .