Math problem! need help fast!!

theiswe:

f(x) = 3^((1/3)(x^3 - 3x)) = 0 Df = R (it says that, what does it mean?)

changed it to this:
f(x) = 3^(x^3/3 - x) -1 = 0

Answer:
x1: (0,0)
x2: (sqrt3, 0)
x3: (-sqrt3, 0)

Tried to solve this by hand. Found (0,0) alright, but cant seem to find how X can be +- square root 3..

PLEASE help fast!! ))

It is getting pretty late here, so I'm going to have to go, but I'll bump this up so it will be answered.

theiswe:

f(x) = 3^((1/3)(x^3 - 3x)) = 0 Df = R (it says that, what does it mean?)

changed it to this:
f(x) = 3^(x^3/3 - x) -1 = 0

Answer:
x1: (0,0)
x2: (sqrt3, 0)
x3: (-sqrt3, 0)

Tried to solve this by hand. Found (0,0) alright, but cant seem to find how X can be +- square root 3..

PLEASE help fast!! ))

What you have here is 3 ^{1/3(x^3-3x)} = 0

Now 3 to the power of anything can never = 0 therefore there are no solutions.
3 to the power of anything is greater than 0.

Now Df = R is not standard maths notation. Not that i have ever seen anyway.
I think it means The domain of the function is all real numbers. (in other words, x can = any real number)
f(x) never equals zero.

theiswe:

f(x) = 3^((1/3)(x^3 - 3x)) = 0 Df = R (it says that, what does it mean?)

changed it to this:
f(x) = 3^(x^3/3 - x) -1 = 0

Answer:
x1: (0,0)
x2: (sqrt3, 0)
x3: (-sqrt3, 0)

Tried to solve this by hand. Found (0,0) alright, but cant seem to find how X can be +- square root 3..

PLEASE help fast!! ))

Hi again theiswe,
I only just noticed the 2nd bit out your question, was that there before?
anyway, lets have another look.

f(x) = 3^((x^3)/3 - x) -1 = 0
so what we have now is
3 ^{((x^3)/3 - x)} -1 = 0
3 ^{((x^3)/3 - x)} = 1
3 ^{((x^3)/3 - x)} = 3 ⁰

((x^3)/3 - x) = 0
x ³/3 - x = 0
Multiply both sides by 3
x ³ - 3x = 0
x(x ² - 3 ) = 0
x (x- sqrt3) (x+ sqrt3) = 0
so x = 0, sqrt3 or -sqrt3