Evaluate 1/((y+3)^2-9)

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

Step 1

Simplify the denominator.

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

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

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

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

Simplify.

Add $3$ and $3$ .

$\frac{1}{(y + 6) (y + 3 - 3)}$

Subtract $3$ from $3$ .

$\frac{1}{(y + 6) (y + 0)}$

Add $y$ and $0$ .

$\frac{1}{(y + 6) y}$

Step 2

Reorder factors in $\frac{1}{(y + 6) y}$ .

$\frac{1}{y (y + 6)}$

Evaluate 1/((y+3)^2-9)

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