Question

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

Answer 1

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`