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$(-1)^3 = (-1)^{6/2} = ((-1)^6)^{1/2} = 1^{1/2} = 1$

So it comes $(-1)^3 = 1$

can anybody explain where exactly the mistake in calculation?

Chiranjeev_Kumar
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Ajay Sharma
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    $\sqrt 1=\pm 1$ – Chiranjeev_Kumar Aug 22 '15 at 17:08
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    You're using the wrong logarithm of $1$ to compute $1^{1/2}$. You used a value of $2k\pi i$ with an even $k$, but since it arose from $(-1)^6$, you must take one of the form $(4m+2)\pi i$ for the formula $(a^b)^c = a^{bc}$ to hold. – Daniel Fischer Aug 22 '15 at 17:10
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    Many questions with the telling tag fake-proofs depend on "conveniently" forgetting that (complex) powers with a fractional exponent are multivalued, so assuming that usual rules apply to them leads to ludicrous identities. See also this common variant. – Jyrki Lahtonen Aug 22 '15 at 17:14

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It is only generally true that $(a^b)^c=a^{bc}$ for $a$ a positive real, unless both $b,c$ are integers.

It gets worse with complex numbers and exponents.

Exponentiation is most naturally seen as a multi-valued function. Then one of the values of $((-1)^6)^{1/2}$ is equal to $(-1)^3$.

In general, every value of $a^{bc}$ will be one value of $(a^b)^c$, but not necessarily visa versa.

Thomas Andrews
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