Simplify (y^3-3y^2+y-3)/(y^2-9)

$\frac{y^{3} - 3 y^{2} + y - 3}{y^{2} - 9}$

Simplify the numerator.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(y^{3} - 3 y^{2}) + y - 3}{y^{2} - 9}$

Factor out the greatest common factor (GCF) from each group.

$\frac{y^{2} (y - 3) + 1 (y - 3)}{y^{2} - 9}$

Factor the polynomial by factoring out the greatest common factor, $y - 3$ .

$\frac{(y - 3) (y^{2} + 1)}{y^{2} - 9}$

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 3$ .

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

Cancel the common factor of $y - 3$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{y^{2} + 1}{y + 3}$