Answer
$6y^{n+1}-12y$
Work Step by Step
Factor the numerators and the denominators, when needed to have:
$\\=\dfrac{(y^{2n}-4)(y^{2n}+4)}{y^{2n}+4} \cdot \dfrac{6y}{y^n+2}
\\=\dfrac{(y^{n}-2)(y^n+2)(y^{2n}+4)}{y^{2n}+4} \cdot \dfrac{6y}{y^n+2}$
Cancel the common factors to have:
$\require{cancel}\\=\dfrac{(y^{n}-2)\cancel{(y^n+2)}\cancel{(y^{2n}+4)}}{\cancel{y^{2n}+4}} \cdot \dfrac{6y}{\cancel{y^n+2}}
\\=6y(y^n-2)
\\=6y^{n+1}-12y$
