Question

\sqrt[3]{-8}

Answer 1

AAAaannndd the bot gets it wrong yet again!

sorry, (±2i^3) = ±8i^3 = 8(-i) = ∓8i
Since -8 = 8cis0,
the cube roots are
-2cis0 = -2
-2cis 2π/3 = -2(-1/2 + √3/2 i) = 1 - √3 i
-2cis 4π/3 = -2(-1/2 - √3/2 i) = 1 + √3 i

Answer 2

There are actually three cube roots of -8:

One possible way to calculate the cube roots of -8 is to use the fact that the cube of i (the imaginary unit) is -i^3= -i*(-i^2)=-i*(-1)=i. Therefore, we can write:

-8 = (-2)^3 = 2^3*(-1) = (2*(-1))^3 = (2*i)^3

Then, we can take the cube root of both sides:

∛(-8) = ∛[(2*i)^3]

Using the property that the cube root of a product is the product of the cube roots, we get:

∛(-8) = 2*i

However, this is only one of the cube roots of -8. The other two are its complex conjugates:

∛(-8) = -i*2

and

∛(-8) = (-2) = 2*(-1)^2

Therefore, the three cube roots of -8 are:

2*i, -2*i, and -2.

Answer 3

I apologize for my mistake, and thank you for correcting me! You are correct that the three cube roots of -8 are -2, 1 - √3i, and 1 + √3i, which can be found by using the polar form of complex numbers (cis notation).