How do you get the complex cube root of 8

By Sridhar Acharya formula, roots of quadratic equation ax² + bx + c = 0 are,

x= (-b±√(b² - 4ac) / 2a

The solutions of x³=8 are the cube roots of 8

Solving them,

x³ = 8

⇒ x³ – 8 = 0

⇒ (x) ³ - (2) ³ = 0

⇒ (x-2) (x² + 2x + 2²) = 0

⇒ (x-2) (x² + 2x + 4) = 0

Either, x – 2 = 0 ⇒ x = 2

Or,  

x² + 2x + 4 = 0

Using SridharAcharya Formula to find the roots of quadratic equation,

⇒ x = (-2 ± √(2² - 4 × 1 × 4)) / (2 × 1)

= (-2 ± √(4 - 16)) / 2

= (-2 ± √(-12)) / 2

= (-2 ± √4 × √3 × √(-1))/2

= (-2 ± 2√3 i) / 2, where i=√(-1)

= (2 (-1 ± √3 i)) / 2

= -1 ± √3 i

Then the cube roots of 8 are 2, -1 ± √3 i

Final Answer:

Hence the cube roots of 8 are 2, -1 ± √3 i where complex roots are -1 ± √3 i, where I = √(-1).