What is the exact value of this expression? $ \left ( -2 \sqrt 2 \right )^{2 \over 3} $ Isn't $2$ one of the answer? Wolfram gets $-1+i \sqrt 3$. Is root multivariable function for complex number?
How to evaluate $(-2\sqrt2)^{2/3}$?
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algebra-precalculus
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0$(-2\sqrt2)^{2/3}=2$. See http://en.wikipedia.org/wiki/Exponentiation#Rational_exponents and http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities – 2012-08-23
1 Answers
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Let $x = (-2 \sqrt 2)^{2 \over 3}$.
$x^3 = (-2\sqrt{2})^2 = 8$.
Let $\omega \neq 1$ be a root of $x^3 = 1$. Then, the roots of $x^3 = 8$ are $2, 2\omega, 2\omega^2$. Let's compute $\omega$.
$x^3 - 1 = (x - 1)(x^2 + x + 1)$.
Hence $\omega = \frac{-1 + i\sqrt{3}}{2}$ or $\frac{-1 - i\sqrt{3}}{2}$.
Hence the roots of $x^3 = 8$ are $2, -1 + i\sqrt{3}, -1 - i\sqrt{3}$.
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0@$M$onkeyD.Luffy , any introductory text book in complex numbers must include this stuff. Try college algebra or stuff like that (though in some parts, like here, this is covered in high school), or simply google "complex numbers, roots of unit, roots of complex numbers " or something like that. There are millions of sites. – 2012-08-24