## Intermediate Algebra: Connecting Concepts through Application

$(a+3b)(a-3b)(a^4+9a^2b^2+81b^4)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $a^6-729b^6 ,$ use the factoring of the sum or difference of $2$ cubes. Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $a^6$ and $729b^6$ are both perfect cubes (the cube root is exact). Hence, $a^6-729b^6$ is a $\text{ difference }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes, which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$, the expression above is equivalent to \begin{array}{l}\require{cancel} (a^2)^3-(9b^2)^3 \\\\= (a^2-9b^2)[(a^2)^2+a^2(9b^2)+(9b^2)^2] \\\\= (a^2-9b^2)(a^4+9a^2b^2+81b^4) .\end{array} The expressions $a^2$ and $9b^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $a^2-9b^2 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares, which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} [(a)^2-(3b)^2](a^4+9a^2b^2+81b^4) \\\\= (a+3b)(a-3b)(a^4+9a^2b^2+81b^4) .\end{array}