How would you solve a negative number raised to a fraction a/b if b is odd and a is evem? Ignoring imaginary numbers
i.e $(-1)^\frac23$ Calculator returns an error
$(-1)^\frac 13 (-1)^\frac 13$ = -1.-1 = 1 (By law of indices)
or
$(-1^\frac13 )^2$ = 1
or
$(-1^2)^\frac13$ = 1
What about for other cases of a and b?
For fractionals with something other than 1 in the numerator, simply use the law of indices which you noticed yourself:
$$a^{b/c}=\left(a^{b}\right)^{1/c}$$
The more interesting part of this question is the denominator.
Remember that $a^{1/b}=c$ means that $c^b=a$, solving for positive $a$ is always possible, however when $a$ is negative you have to notice the basic laws of what happens to the sign of the number on multiplication.
The thing is that if you have a negative number $c$ then $c^2$ is positive since negative times negative is positive. You then see that $c^3$ is negative since $c^3=c^2\cdot c$ and $c^2$ is positive and $c$ is negative.
Using this you can conclude that solving when the base is negative depends on whether the number in the exponent is even or odd.
The cases are these:
The exponent has an even denominator
Since all numbers to an even power are positive, there simply are no solutions and it's undefined.
The exponent hass an odd denominator
Negative numbers to an odd power are negative, so solving this is simply negative the solution of the equation with a positive base.
Your example pretty much handles the generic case, as long as you are dealing strictly with real numbers.
If $b$ is odd and $a$ is even, you would be extracting an odd root (which always exists, whether the base is positive or negative), and raising the result to an even power (which gets rid of the negative sign).
The same thing happens if you do it in the reverse order. So the result would be a well-defined positive number.
The real problems arise if $b$ is even. In that case, you will have to insist on a set of ground rules (for example, you must insist that $a/b$ is reduced to lowest terms, otherwise you would have a problem determining $(-1)^{2/6}$ which should be the same as $(-1)^{1/3}$).
