Find the Domain 1/(x^2-64)-7/(x-8)=1/(x+8)

$\frac{1}{x^{2} - 64} - \frac{7}{x - 8} = \frac{1}{x + 8}$

Step 1

Set the denominator in $\frac{1}{x^{2} - 64}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 64 = 0$

Step 2

Solve for $x$ .

Add $64$ to both sides of the equation.

$x^{2} = 64$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{64}$

Simplify $\pm \sqrt{64}$ .

Rewrite $64$ as $8^{2}$ .

$x = \pm \sqrt{8^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 8$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 8$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 8$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 8, - 8$

Step 3

Set the denominator in $\frac{7}{x - 8}$ equal to $0$ to find where the expression is undefined.

$x - 8 = 0$

Step 4

Add $8$ to both sides of the equation.

$x = 8$

Step 5

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 8) \cup (- 8, 8) \cup (8, \infty)$

Set-Builder Notation:

${x | x \neq 8, - 8}$

Step 6