What is the 6th root of a negative 64 i.e.(-64)?

Question

Who can help me with the answer to this, please? Thanks!

Answer 1

There are 6 of them (there are 6 6th roots of every number, and in general, there are n nth roots of every number)

So far some of the others have identified one of them, 2i.
-2i is another
and also:
√3 + i
√3 - i
-√3+ i
-√3 - i

Wal C has a nice derivation of this....
using a + bi = z = e^(iθ) = cos(θ) + i*sin(θ)

Answer 2

THER IS NO REAL ROOT i.e. u cant have a number that when multiplied by itslef that gives you a negative number.
the square r. of 64 is 8 but not -64.

theres another way tho
theres an imaginary number called i (the letter i ) it represents the square root of -1.
There fore you can rewrite the q. as ' the square root of 64 times -1
which is the same thing as -64.
Since the sq of -1 is i ... therefore i gets out of the square root thingi.
next the 64 gets out as 8

therefore the answer is 8i

Answer 3

Since 2x2x2x2x2x2 = 64, this root is imaginary and would be 2i

Answer 4

This is a complex number question

Let z = e^iθ
Thus z^6 = e^6iθ = -64 = 64e^(±iπ), 64e^(±i3π), 64e^(±i5π)
(as there are 6 roots in the range -π ≤ θ ≤ π equally spaced around a circle with radius 64^(1/6) ie 2)

So z = 2e^(±iπ/6), 2e^(±i3π/6), 2e^(±i5π/6)
= 2(cos(π/6) + isin(π/6)),
2(cos(-π/6) + isin(-π/6)),
2(cos(π/2) + isin(π/2)),
2(cos(-π/2) + isin(-π/2)),
2(cos(5π/6) + isin(5π/6)),
2(cos(-5π/6) + isin(-5π/6))

ie z = (-64)^(1/6) = √3 ± i, ±2i, -√3 ± i

Answer 5

If i^2 = -1, then i^6 = -1. I'm guessing that is all you need to figure out the rest. :)

Answer 6

2i^6 = -64
[ (2i)^2 ]^3
[ (2i * 2i) ]^3
(-4)^3
-64

Answer 7

+/-2 if it is 64
x^6=-64
x^2=-1
x=+/-i