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.