cube root of -64?

Answer 1

The most obvious answer is cube root of -64 is -4:
-4 * -4 * -4 = -64

However, there are actually *three* cube roots to -64, one real (-4) and two imaginary.

To figure the complex (imaginary) roots, start with:
x^3 = -64

Change this to:
x^3 + 64 = 0

We know one root is -4, so factor out (x + 4). (You can use synthetic division for this part):

(x + 4)(x^2 -4x + 16) = 0

One solution is -4, as we knew before:

The other solutions are obtained by applying the quadratic formula to:

x^2 - 4x + 16 = 0

This gives :

x = {4 ± sqrt[(-4)^2 - 4(1)(16)]} / (2 * 1)

= [4 ± sqrt(-48)] / 2
= [4 ± 4 sqrt(-3)] / 2

= 2 ± 2 sqrt(-3)

= 2 ± 2 sqrt(3) i

So your *three* cube roots of (-64) are:
2 + 2 sqrt(3) i
2 - 2 sqrt(3) i
-4

You can confirm that by cubing each of the answers. I've given you links to Google calculator with the results.

Answer 2

Since 4*4*4=64, and when you have a negative to a odd n power, you get a negative, you are able to take the n root of that number. The cube root of 64 is 4, therefore, the cube root of -64 is -4.

Answer 3

it depends on your unit vectors.

If they are 1, then the cube root of -64 is -4
n^3=-64

factors of -64 are -1 4 4 4
-1^3=-1

If the unit vector is negative one then the cube root is four.

If the unit vector is the square root of negative one, then your cube root is 0-4*i

Answer 4

cube root for -64 is -4

Answer 5

Its -4 because -4 * -4 is 16 * -4 is -64

Answer 6

I only know 2: 0+2i and 0-2i. The other four, have fun trying to figure out.

Oh, the CUBE root! In that case, it's -4.

Answer 7

it has got 3 cube roots one real and 2 complex

they are -4, -4 w, -4 w^2

Answer 8

-4x-4x-4 = -64
Therefore, the cube root of -64 is -4

Answer 9

-4


64 = -4 * -4 * -4

Answer 10

4.