$ (-1)^{3}=((-1)^2)^\frac {3}{2}=(1)^\frac {3}{2}=1$
See, I know it's wrong but don't know why . Which rule is disobeyed here? And which rule should be followed in these cases?
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$(a^{p})^{q}=a^{pq}$ is not valid for negaitve $a$. – Kavi Rama Murthy Mar 24 '22 at 12:25
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1The punchline is that $a^{bc}$ is not necessarily equal to $(a^b)^c$ when non-positive real numbers are involved as the base, i.e. your first equals sign was incorrect. – JMoravitz Mar 24 '22 at 12:25
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The exponent multiplication rule does not hold in the complex case. In general for a complex number $z \in \mathbb{C}$ and real numbers $a,b \in \mathbb{R}$,
$$(z^a)^b \ne z^{ab}$$
Of course you can find some special cases where it does hold, but generally it does not. You can see some examples here and a proof here.
Now while it doesn't look like you're dealing with complex numbers, implicitely you are, since
$$(-1)^{1/2} = \sqrt{-1} = i$$
So immediately whenever there are negative values raised to a non-whole value, the complex exponentiation rules apply.
Daniel P
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