Solve 8x^3+27=0?

Answer 1

NO you solve it.

Answer 2

8x^3+27=0
8x^3=-27
x^3=-27/8
x = cubic_root(-27/8)
You know that n_root operation produces n values. For getting them we have to use complex numbers. If you don't know them yet you can only get one root (the real one) which is -3/2. If you already know them you should notice the 3 roots are:

3/2[cos(Pi+0*2*Pi/3)+i*sin (Pi+0*2*Pi/3)]
=-3/2=-0.75
,
3/2[cos(Pi+1*2*Pi/3)+i*sin (Pi+1*2*Pi/3)]
= 0.75-1.2990*i
and
3/2[cos(Pi+2*2*Pi/3)+i*sin (Pi+2*2*Pi/3)]
= 0.75+1.2990*i

Another way is factoring 8x^3+27 with (x-3/2), which is one of the known products. You get:
8x^3+27=(x+3/2)*(8*x^2-12*x+18)
you solve now (8*x^2-12*x+18):

x=-(3*sqrt(3)*i-3)/4 = 0.75-1.2990*i
x=(3*sqrt(3)*i+3)/4 = 0.75+1.2990*i

Answer 3

x^3 = -27/8

x = - 3/2

other roots are complex

8x^3+27=0

factorising
(2x+3)(4x^2 -6x + 9) = 0

solving
(4x^2 -6x + 9) = 0
using quadratic formula you will get the complex roots

Answer 4

x = -(3/2)

Answer 5

8x^3 + 27 = 0
8x^3 = −27
x^3 = −3.375
x = −1.5

Answer 6

8x. x= -30
cant sq root - 30/8