Answer
$\left(9y+\dfrac{1}{7}\right)\left(9y-\dfrac{1}{7}\right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
81y^2-\dfrac{1}{49}
,$ use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The expressions $
81y^2
$ and $
\dfrac{1}{49}
$ are both perfect squares and are separated by a minus sign. Hence, $
81y^2-\dfrac{1}{49}
,$ 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}
\\\\=(9y)^2-(\frac{1}{7})^2
\\\\=
\left(9y+\dfrac{1}{7}\right)\left(9y-\dfrac{1}{7}\right)
.\end{array}
