Determine the cube root of 27(cos(7pi/6)+i sin (7pi/6)).There is more then one answer.

Question

Determine the cube root of 27(cos(7pi/6)+i sin (7pi/6)). There is more then one answer.

Answer 1

We can also find the cube roots of 27 algebraically.

 

A cube root of 27 is a number, x, such that x³ = 27.

 

So, x³  - 27 = 0

 

Using the formula for the difference of cubes, we have:

 

(x-3)(x² + 3x + 9) = 0

 

x = 3 or x = [-3 ± √(9 - 36)]/2

 

Simplifying, we get x = 3 or x = (-3 ± 3√3 i)/2

 

Answer 2

Whoever voted against Mark M's solution (the Mark M from Carson, CA) was being grossly unfair.  At the point in a Trig course where DeMoivre's theorem is being discussed, this is the natural way to resolve it (exponent 1/3 is applied to the modulus and1/3  is applied to the θ as a multiplicative factor.  Then other roots simply increase argument θ/3 by 2pi / 3 to get the second & third of the three complex cubic roots.

 

However, in his 2nd & 3rd lines, the parentheses following pi should be moved to the right to follow 18 in each instance.

Answer 3

Using DeMoivre's Theorem three cube roots exist.

1) 27^1/3(cos 7π/18 + i sin 7π/18)

2) 27^1/3(cos (19π)/18 + i sin (19π)/18)

3) 27^1/3(cos (31π)/18 + i sin (31π)/18)