$$\frac{\left(2y-3\right)\left(y^{2}-1\right)}{\left(y+1\right)\times 12}\times \frac{8}{2y^{2}-y-3}$$

$$\frac{\left(2y-3\right)\left(y^{2}-1\right)\times 8}{\left(y+1\right)\times 12\left(2y^{2}-y-3\right)}$$

$$\frac{2\left(2y-3\right)\left(y^{2}-1\right)}{3\left(y+1\right)\left(2y^{2}-y-3\right)}$$

$$\frac{2\left(y-1\right)\left(2y-3\right)\left(y+1\right)}{3\left(2y-3\right)\left(y+1\right)^{2}}$$

$$\frac{2\left(y-1\right)}{3\left(y+1\right)}$$

$$\frac{2y-2}{3y+3}$$

$$\frac{\left(2y-3\right)\left(y^{2}-1\right)}{\left(y+1\right)\times 12}\times \frac{8}{2y^{2}-y-3}$$

$$\frac{\left(2y-3\right)\left(y^{2}-1\right)\times 8}{\left(y+1\right)\times 12\left(2y^{2}-y-3\right)}$$

$$\frac{2\left(2y-3\right)\left(y^{2}-1\right)}{3\left(y+1\right)\left(2y^{2}-y-3\right)}$$

$$\frac{2\left(y-1\right)\left(2y-3\right)\left(y+1\right)}{3\left(2y-3\right)\left(y+1\right)^{2}}$$

$$\frac{2\left(y-1\right)}{3\left(y+1\right)}$$

$$\frac{2y-2}{3y+3}$$