Q 268 - Intermediate Algebra

Question

In the following exercises, factor completely. 

$y^6+1$

Step-by-Step Solution

Answer

The expression is factored as $y^6+1=(y^2+1)(y^4-y^2+1)$

Step 1. Given Information

Consider the expression $y^6+1$

The objective is to factor the expression completely.

Step 2. Factor using the sum of cubes

The sum of cubes can be factored as $a^3+b^3=(a+b)(a^2-ab+b^2)$

In the given binomial $y^6+1$, the first term $y^6$ is the cube of $y^2$ and the second term $1$ is the cube of $1$. So it can be factored as

$\begin{array}{rcl}{y}^6+1& =& {\left(y^2\right)}^3+{\left(1\right)}^3 \\ & =& (y^2+1)\{{\left(y^2\right)}^2-y^2·1+{\left(1\right)}^2\} \\ & =& (y^2+1)(y^4-y^2+1)\end{array}$

Step 3. Check the factors

Multiply the factors to check the solution  

$\begin{array}{rcl}(y^2+1)(y^4-y^2+1)& =& y^2(y^4-y^2+1)+1(y^4-y^2+1)\\ & =& y^6-y^4+y^2+y^4-y^2+1\\ & =& y^6+1\end{array}$

And we get the given expression. So the expression is correctly factored.