The number 1 has three distinct cube roots in the complex number system. These roots are 1, (-\frac{1}{2} + \frac{\sqrt{3}}{2}i), and (-\frac{1}{2} - \frac{\sqrt{3}}{2}i). In polar form, these roots can be represented as (1), (1 \text{cis} \frac{2\pi}{3}), and (1 \text{cis} \frac{4\pi}{3}), where "cis" is shorthand for (\cos + i\sin).
4
All numbers have cube roots (not necessarily integral cube roots) so every prime has cube roots.
The real cube root of 1 is 1, since 13 = 1. There also a pair of complex cube roots.
Yes.
The number -512 has three cube roots in the complex number system. In general, any non-zero complex number has three distinct cube roots. For -512, these roots can be expressed in the form ( r^{1/3} (\cos(\theta/3) + i \sin(\theta/3)) ), where ( r ) is the magnitude and ( \theta ) is the argument of the complex number. The three cube roots are evenly distributed around the unit circle in the complex plane.
1
there is no cube roots in negative
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All numbers have cube roots (not necessarily integral cube roots) so every prime has cube roots.
The real cube root of 1 is 1, since 13 = 1. There also a pair of complex cube roots.
Yes.
3√0.125 = 0.5 (there is only 1 real cube root of 0.125).
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-21 + 1.7320508i1 - 1.7320508i
There are 3 cube roots of 27. There are 2 square roots of 27 ( or any real number ). There are 4 fourth roots of 27 and so on:)
The ancients - Egyptians or Greeks. They probably came across the square root of 2 when considering the diagonal of a square with sides of length 1. The cube root of 3 would have arisen, similarly, with the principal diagonal of a unit cube.
Some calculators don't have cube roots, so you could use X to the power of 1/root. x^(1/root). The cube root of 64 equals 4 because 4*4*4 (or 4^3) gives 64. You can reverse these powers by reciprocating the exponent: 64^(1/3) = 4