Answer
$y=\{ \pm3i, 1 \}$
Work Step by Step
Using factoring by grouping, the given equation, $
5y^3+45y-5y^2-45=0
,$ is equivalent to
\begin{array}{l}\require{cancel}
(5y^3+45y)-(5y^2+45)=0
\\\\
5y(y^2+9)-5(y^2+9)=0
\\\\
(y^2+9)(5y-5)=0
.\end{array}
Equating each factor to zero, then,
\begin{array}{l}\require{cancel}
y^2+9=0
\\\\
y^2=-9
\\\\
y=\pm\sqrt{-9}
\\\\
y=\pm\sqrt{-1}\cdot\sqrt{9}
\\\\
y=\pm i\cdot3
\\\\
y=\pm 3i
,\\\\\text{OR}\\\\
5y-5=0
\\\\
5y=5
\\\\
y=\dfrac{5}{5}
\\\\
y=1
.\end{array}
Hence, the solution set is $
y=\{ \pm3i, 1 \}
.$
