Simplify (y^3-2y^2-3y)/(y^3+1)

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

Step 1

Simplify the numerator.

Factor $y$ out of $y^{3} - 2 y^{2} - 3 y$ .

Factor $y$ out of $y^{3}$ .

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

Factor $y$ out of $- 2 y^{2}$ .

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

Factor $y$ out of $- 3 y$ .

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

Factor $y$ out of $y \cdot y^{2} + y (- 2 y)$ .

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

Factor $y$ out of $y (y^{2} - 2 y) + y \cdot - 3$ .

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

Factor $y^{2} - 2 y - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

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

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

Step 2

Simplify the denominator.

Rewrite $1$ as $1^{3}$ .

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

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = y$ and $b = 1$ .

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

Simplify.

Multiply $- 1$ by $1$ .

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

One to any power is one.

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

Step 3

Cancel the common factor of $y + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify (y^3-2y^2-3y)/(y^3+1)

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