If #root(3)(3(root(3)x - 1/(root(3)x))) = 2#, then #root(3)x + 1/(root(3)x) =# what?

We start with the original function:
#root(3)(3(root(3)x - 1/(root(3)x))) = 2#

and we want to solve for:

#root(3)x + 1/(root(3)x)# (note the + instead of -)

which means we need to do a bit more work and solve for #x# and then use our x value to solve for:

#(root(3)x + 1/(root(3)x))#

We then take each side to the exponent 3 (which gets rid of the cube root on the left hand side):
#(root(3)(3(root(3)x - 1/(root(3)x))))^3 = 2^3#

Taking the cube root of a number is equivalent to taking a fractional exponent of 1/3 and so this simplifies to:
#(3(root(3)x - 1/(root(3)x)))^(3/3) = 8#

#3(root(3)x - 1/(root(3)x)) = 8#

We then divide both sides by 3 (to get ride of the multiplication by 3 on the left hand side)
#3(root(3)x - 1/(root(3)x))/3 = 8/3#

#root(3)x - 1/(root(3)x) = 8/3#

Next we multiply both sides by #root(3)x# to remove it from the denominator of the second term

#root(3)x^2 - 1 = (8*root(3)x)/3#

#root(3)x^2 - (8*root(3)x)/3 - 1 = 0#

You might not immediately recognize it, but this is a quadratic equation. To make that more obvious, we're going to temporarily substitute #root(3)x#. Let's choose a new variable #y# to represent #root(3)x#:

#y = root(3)x#

And now our equation becomes:

#y^2 - 8y/3 - 1 = 0#

which more closely resembled what we expect from a quadratic equation. Using the quadratic formula and the letter #y# now instead of the traditional #x#:

# y = (-b ±sqrt(b^2 - 4ac))/(2a) #

we find that:

#y = 3# and #y=-1/3#

Now we'll substitute back our function of x for y: #y = root(3)x#

#root(3)x = 3#
#x = 3^3#
#x = 27#

and

#root(3)x = -1/3#
#x = (-1/3)^3#
#x = -1/27#

So we have two solutions for x:
#x = 27# and #x=-1/27#

Using either of these values (#27 or -1/27#), we should be able to find the answer to the original problem:

#root(3)x + 1/(root(3)x)#
#= root(3)27 + 1/(root(3)27)#

We know that #root(3)27 = 3# which means:

#= 3 + 1/3#

#= (9+1)/3 #

#= 10/3 #