Find all three distinct roots for X³ =1. (two of them are complex mumber). And show that sum of all roots is zero.​

x ³ = 1

x ³ - 1 = 0

x ³ - 1³ = 0

Factorize the right side as a difference of cubes:

(x - 1) (x ² + x + 1) = 0

Then

x - 1 = 0   or   x ² + x + 1 = 0

The first equation yields x = 1 as a root.

For the other equation, rearrange and complete the square to get the other two roots,

x ² + x + 1/4 = -3/4

(x + 1/2)² = -3/4

x + 1/2 = ±√(-3/4) = ± i √3/2

x = -1/2 ± i √3/2

x = (-1 ± i √3)/2

Since complex roots occurs in conjugate pairs, taking the sum of these two roots eliminates the imaginary part:

(-1 + i √3)/2 + (-1 - i √3)/2 = (-1 - 1)/2 = -2/2 = -1

and adding to 1 indeed shows that the sum of roots is zero.

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)