\(\bf \cfrac{y^2-5}{y^4-81}+\cfrac{4}{y^4-81}\implies \cfrac{y^2-5+4}{y^4-81}\implies \cfrac{y^2-1}{y^4-81}\\\\
-----------------------------\\\\
\textit{now recall your }\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\\\\
\textit{and keep in mind that }
\begin{cases}
1^2=1\\
1^3=1\\
1^{1,000,000}=1\\
----------\\
81=9^2\\
9=3^2
\end{cases}\\\\\)
\(\bf -----------------------------\\\\
thus
\\\\\\
\cfrac{y^2-1^2}{(y^2)^2-9^2}\implies \cfrac{(y-1)(y+1)}{(\underline{y^2-9})(y^2+9)}
\\\\\\
\cfrac{(y-1)(y+1)}{\underline{(y-3)(y+3)}(y^2+9)}\)
would be a factored version of it
