Find three cube roots for the following complex number. Leave your answers in trigonometric form. (Enter your answers from smallest to largest angle measure.)8(cos 120° + i sin 120°)

To find the cube roots of a complex number in trigonometric form, we use De Moivre's Theorem. The formula for finding the nth roots of a complex number is:

r^(1/n)[cos((θ+2πk)/n) + i sin((θ+2πk)/n)]

where r is the modulus of the complex number, θ is the argument of the complex number, n is the root you are trying to find (in this case, 3 for cube root), and k is an integer from 0 to n-1.

Given the complex number 8(cos 120° + i sin 120°), we have:

r = 8 θ = 120° n = 3

We can find the three cube roots by plugging in k = 0, 1, 2 into the formula:

For k = 0: root1 = 8^(1/3)[cos((120°+2π0)/3) + i sin((120°+2π0)/3)] root1 = 2[cos(40°) + i sin(40°)]

For k = 1: root2 = 8^(1/3)[cos((120°+2π1)/3) + i sin((120°+2π1)/3)] root2 = 2[cos(160°) + i sin(160°)]

For k = 2: root3 = 8^(1/3)[cos((120°+2π2)/3) + i sin((120°+2π2)/3)] root3 = 2[cos(280°) + i sin(280°)]

So, the three cube roots of 8(cos 120° + i sin 120°) are 2[cos(40°) + i sin(40°)], 2[cos(160°) + i sin(160°)], and 2[cos(280°) + i sin(280°)].