#### Answer

domain : $\{\ x \ |\ \ x\neq-8, \ x\neq 8 \ \}$
interval notation: $(-\infty,-8)\cup(-8,8)\cup(8,\infty)$

#### Work Step by Step

See example 1.
Division by 0 is undefined, so
$h(x)$ will be defined for all real numbers except for
the number(s) x that yield zero in the denominator:
$ x^{2}-64=0\qquad$... recognize a difference of squares
$(x-8)(x+8)=0$
$x=8$ or $x=-8\qquad $(zero product principle)
domain : $\{\ x \ |\ \ x\neq-8, \ x\neq 8 \ \}$
interval notation: $(-\infty,-8)\cup(-8,8)\cup(8,\infty)$