Chapter 6.5, Problem 3ACYP

To determine

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

Answer to Problem 3ACYP

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

Explanation of Solution

Given:

The given polynomial is $a^6+b^6$ .

Calculation:

The given polynomial is $a^6+b^6$ .

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

  $\begin{array}{l}{a}^6+b^6\left\{\text{given polynomial}\right\}\\ =\left(a^6-b^6\right)\left\{\text{factor out the GCF}\right\}\\ ={\left(a^3\right)}^2+{\left(b^3\right)}^2\\ \left\{\text{Since both are perfect cubes, so write the factor of the sum of two cubes}\text{.}\right\}\\ \left\{a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\right\}\\ =\left(a^2+b^2\right)\left({\left(a^2\right)}^2-a^2{b}^2+{\left(b^2\right)}^2\right)\\ =\left(a^2+b^2\right)\left(a^4-a^2{b}^2+b^4\right)\end{array}$

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

Hence, the factor of the given polynomial $a^6+b^6=\left(a^2+b^2\right)\left(a^4-a^2{b}^2+b^4\right)$ .