Answer
$\frac{y^n-1}{y^n+4}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{y^{2n}-1}{y^{2n}+3y^n+2}\div\frac{y^{2n}+y^n-12}{y^{2n}-y^n-6}$
Invert the divisor and multiply.
$\Rightarrow \frac{y^{2n}-1}{y^{2n}+3y^n+2}\cdot \frac{y^{2n}-y^n-6}{y^{2n}+y^n-12}$
Let $y^n=x$.
$\Rightarrow \frac{x^{2}-1}{x^{2}+3x+2}\cdot \frac{x^{2}-x-6}{x^{2}+x-12}$
Factor each numerator and denominator as shown below.
$\Rightarrow x^2-1$
$\Rightarrow x^2-1^2$
Use the special formula $a^2-b^2=(a+b)(a-b)$.
$\Rightarrow (x+1)(x-1)$
$\Rightarrow x^2+3x+2$
Rewrite the middle term $3x$ as $2x+1x$.
$\Rightarrow x^2+2x+1x+2$
Group terms.
$\Rightarrow (x^2+2x)+(1x+2)$
Factor each group.
$\Rightarrow x(x+2)+1(x+2)$
Factor out $(x+2)$.
$\Rightarrow (x+2)(x+1)$
$\Rightarrow x^2-x-6$
Rewrite the middle term $-x$ as $-3x+2x$.
$\Rightarrow x^2-3x+2x-6$
Group terms.
$\Rightarrow (x^2-3x)+(2x-6)$
Factor each group.
$\Rightarrow x(x-3)+2(x-3)$
Factor out $(x-3)$.
$\Rightarrow (x-3)(x+2)$
$\Rightarrow x^2+x-12$
Rewrite the middle term $x$ as $4x-3x$.
$\Rightarrow x^2+4x-3x-12$
Group terms.
$\Rightarrow (x^2+4x)+(-3x-12)$
Factor each group.
$\Rightarrow x(x+4)-3(x+4)$
Factor out $(x+4)$.
$\Rightarrow (x+4)(x-3)$
Substitute all the factors into the given expression.
$\Rightarrow \frac{(x+1)(x-1)}{(x+2)(x+1)}\cdot \frac{(x-3)(x+2)}{(x+4)(x-3)}$
Cancel common terms.
$\Rightarrow \frac{x-1}{x+4}$
Back substitute $x=y^n$.
$\Rightarrow \frac{y^n-1}{y^n+4}$.
