Question

Solve 64x^3 +1 =0

the back of the texts books gives me the answers -1/4 and 1 plus or minus i√3 all over 8
I understand where the -1/4 comes from . I just don't understand the second answer. I know they used the quadratic formula to get it but I always end up with something to totally different

Answer 1

Expand the expression using the formula:

`a^3+b^3=(a+b)(a^2 - ab + b^2)`


`64x^3+1=0 \\ (4x)^3+1^3=0 \\ (4x+1)((4x)^2- 4x * 1+1^2)=0 \\ (4x+1)(16x^2-4x+1)=0 \\ 4x+1=0 \ \lor \ 16x^2-4x+1=0`

The first equation:

`4x+1=0 \\ 4x=-1 \\ x=-(1)/(4)`

The second equation:

`16x^2-4x+1=0 \\ \\ a=16 \\ b=-4 \\ c=1 \\ b^2-4ac=(-4)^2-4 * 16 * 1=16-64=-48 \\ √(b^2-4ac)=√(-48)=√(48) * √(-1)=√(16 * 3) * √(-1)=\pm 4√(3) i \\ \\ x=(-b \pm √(b^2-4ac))/(2a)=(-(-4) \pm 4√(3)i)/(2 * 16)=(4 \pm 4√(3)i)/(32)=(4(1 \pm √(3)i))/(4 * 8)=(1 \pm √(3)i)/(8) \\ x=(1-√(3)i)/(8) \ \lor \ x=(1+√(3)i)/(8)`

The final answer:

`\boxed{x=-(1)/(4) \hbox{ or } x=(1-√(3)i)/(8) \hbox{ or } x=(1+√(3)i)/(8)}`