Solve for a a(b^2-4)=n

$a (b^{2} - 4) = n$

Step 1

Divide each term in $a (b^{2} - 4) = n$ by $b^{2} - 4$ .

$\frac{a (b^{2} - 4)}{b^{2} - 4} = \frac{n}{b^{2} - 4}$

Step 2

Simplify the left side.

Cancel the common factor of $b^{2} - 4$ .

Cancel the common factor.

$\frac{a (b^{2} - 4)}{b^{2} - 4} = \frac{n}{b^{2} - 4}$

Divide $a$ by $1$ .

$a = \frac{n}{b^{2} - 4}$

Step 3

Simplify the right side.

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$a = \frac{n}{b^{2} - 2^{2}}$

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

$a = \frac{n}{(b + 2) (b - 2)}$