Where is the step that caused the extraneous solution?

An even root cannot have negative values so the solution $\sqrt[6]{x}=-2$ is not valid in $\mathbb{R}$

There is no real solution for $x$ in $x^{1/6}=-2.$ When you took the $6$th power of both sides, you lost information about the sign.

I'll take the algebraist's approach for this one.

The general rule is something like this: in an equation $L = R$ where $L$, $R$ are arbitrary terms, there are two general kinds of steps:

1. Rearraging one side

If you have an equation $$ L = R $$ then you can turn it into $$ L' = R $$ as long as $L$ and $L'$ are equivalent. You can do the same on the right-hand side, of course. This covers things like multiplying out brackets or taking a common denominator for fractions.

In your case, rewriting $x^{1/3} + x^{1/6} - 2$ as $(x^{1/6} + 2)(x^{1/6} - 1)$ is such a step.

This kind of step is always fine.

2. Applying a function $f$ to both sides.

Imagine you have an equation $x - 2 = 3$. Normally you'd "add 2 on both sides", what you're actually doing is applying the function $f(T) = T + 2$ on both sides (*1):

$$ \begin{array}{rrcl} & x - 2 & =& 3 \\ \Rightarrow & f(x-2) &=& f(3) \\ \Rightarrow & x - 2 + 2 &=& 3 + 2 \\ \Rightarrow & x &=& 5 \end{array} $$

The last step is a "rearranging" step on both sides.

The two most common examples of this case are $f(T) = T + C$ and $f(T) = TC$ where $C$ is any number (except $0$, in the second case), which corresponds to adding/subtracting and multiplying with a number to simplify the equation.

We have several cases for this technique:

• If $f$ is bijective (invertible), this kind of step is always fine. $f(T) = T + 2$ is bijective: its inverse is $f^{(-1)}(T) = T - 2$.

• If $f$ is not bijective, the step is still allowed but with caution, as it may cause extra solutions to appear, or lose some solutions (*2).

  The classic example of gaining extra solutions is applying $f(T) = T^2$ to $x = 2$ to get $x^2 = 4$ which has two solutions $2$ and $(-2)$. Conversely, applying $f(T) = \sqrt T$ to $x^2 = 4$ gets you $x = 2$, losing the negative root.

  In your case, applying $f(T) = T^6$ was the non-bijective function.

• (EDIT - adding this to avoid possible confusion) The function that you apply can itself contain the variable $x$. For example in the equation $x + 1 = 2x$ you want to "subtract an $x$ on both sides" with $f(T) = T - x$ to get $1 = x$. This function has the inverse $f^{(-1)}(T) = T + x$, whatever the value of $x$, so this step is fine.

  The place to be careful is when you have something like $$ \frac{1}{x} = \frac{x+1}{x} $$ The obvious function (*3) to choose is $f(T) = Tx$ getting you $1 = x + 1$. If $x$ is not $0$ then this function has an inverse $f^{(-1)}(T) = T/x$, whereas if $x$ turns out to be $0$ - as it does in this case - then you've multiplied by $0$, which is allowed but not very helpful as it gains you an extraneous solution.

  So if your function involves an $x$, you need to check whether it has an inverse for any possible $x$, and if not then check the cases when it doesn't again at the end.

  In this case, if you graph $1/x$ and $(x+1)/x$ you'll see that they don't intersect so there can't be any solutions to this equation; alternatively you could do a rearranging step to get $1/x = 1 + 1/x$ and then either conclude that nothing can ever equal itself plus one, or apply the function $f(T) = T - 1/x$ to get $0 = 1$ which also shows that there are no solutions.

Further notes

There are of course also non-general techniques that you need to solve certain kinds of equations, like splitting $(x-a)(x-b) = 0$ into two equations $x-a=0$ and $x-b=0$.

(*1) Experts in group theory are welcome to moan something about missing a step involving the associative law here.

(*2) To be precise, you can gain or lose solutions if the function is not bijective on the domain you're solving the equation for, in this case the reals. The problem with $\sqrt T$ is further that it's not even defined on all the reals, which is what loses you some solutions.

(*3) Alternatively you could directly choose $f(T) = Tx - 1$ to do "two steps at once". But that's not how they teach it in schools.