Question 29867
h(x) = 3x/ x(x^2-81).
That is h(x) = 3/(x^2-81) (canceling x not zero)
Since division by zero is not defined, h(x) cannot be defined for (x^2-81)=0
And (x^2-81) = 0 
implies (x+9)(x-9) = 0 (using a^2-b^2 = (a+b)(a-b) where a = x and b = 9)
that is for (x+9)=0 giving x=-9 and for (x-9) = 0 giving x = 9
Therefore the funciton h(x) is defined for all x EXCEPTING x = 9 and x = -9
Therefore the domain of the given function is the complement of {-9,9} in the real number system R
That is Domain = R - {-9,9}