Question

How do you solve and check for extraneous solutions in `root3(x-3) =1`?

Answer 1

`x=4` with no extraneous solutions

Explanation:

Given `root(3)(x-3) = 1`

Cube both sides:
`color(white)("XXXX")` `x-3 = 1^3 = 1`

Add 3 to both sides
`color(white)("XXXX")` `x = 4`

Verify by substituting `4` for `x` in the original expression
`color(white)("XXXX")` `root(3)(4-3) = root(3)(1) = 1`
so this solution is not extraneous