Question

What are the real and complex solutions of the polynomial equation? x^3-64=0

Answer 1

Answer:

Step-by-step explanation:

A difference of two perfect cubes,  x^3 - y^3 can be factored into

(x-y) • (x^2 +xy +y^2) =0  equation 1

x ^3 - 64 = 0

(x)^3 - (4)^3 = 0   equation 2

Now substituting equation 2 into equation 1, we get

(x-4) (x^2+(x).(4) +(4)^2) = 0

(x-4) (x^2+4x+16) = 0

so the solutions are

1) x - 4 =0

x=4

x^2 + 4x + 16 = 0

By using the quadratic formula we get the followig solutions:

          - B  ±  √ B2-4AC

 x =   ————————

                     2A

x =(-4-√-48)/2=-2-2i√ 3 = -2.0000-3.4641i

x =(-4+√-48)/2=-2+2i√ 3 = -2.0000+3.4641i