State the domain restrictions.

Question

smurfy14 · Answer

$$x^3 \over \sqrt[3]{1-x^3}$$\]

anonymous · Answer

the denominator cannot be zero, so set 
$$1-x^3=0$$ and solve for x in one step

smurfy14 · Answer

is that the only one? and also theu subscript 3 is not included right?

anonymous · Answer

i am not sure what you mean by "subscript 3"  
the cube root maybe?  if so ignore it since the cubed root of 0 is 0, so you only need to worry about 
$$1-x^3=0$$ or 
$$x=1$$ any other value is good

lgbasallote · Answer

the denominator cannot be zero
the radical cannot be negativec

anonymous · Answer

oh no the radical expression certainly can be negative

anonymous · Answer

it is the cubed root not the square root. if it was square root you cannot have a negative number under the radical.  but for the cubed root that is not a problem

$$\sqrt[3]{-8}=-2$$for example, a perfectly good number

smurfy14 · Answer

so would domain be: (-infinity, 1)?

anonymous · Answer

no the domain would be all real numbers except 1

smurfy14 · Answer

oh ok got it thanks!!!

anonymous · Answer

yw

anonymous · Answer

$$(-\infty,1)\cup(1,\infty)$$