91Ó°ÊÓ

Consider the expression$y^6-1$

The objective is to factor the expression completely.

The difference of squares can be factored as$a^2-b^2=(a-b)(a+b)$

Now for the given binomial $y^6-1$, the first term $y^6$is the square of $y^3$and $1$is the square of $1$. So it can be factored as

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

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

The sum of cubes can be factored as

$a^3+b^3=(a+b)(a^2-ab+b^2)$

So the expression can be further factored as

role="math" localid="1644653795994" $\begin{array}{rcl}(y^3-1)(y^3+1)& =& \{{\left(y\right)}^3-{\left(1\right)}^3\}\{{\left(y\right)}^3+{\left(1\right)}^3\}\\ & =& (y-1)(y^2+y+1)(y+1)(y^2-y+1)\\ & =& (y-1)(y+1)(y^2+y+1)(y^2-y+1)\end{array}$

Multiply the factors to check the solution

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

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