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umka21 [38]
3 years ago
9

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

Mathematics
1 answer:
liraira [26]3 years ago
8 0

<em>x</em> ³ = 1

<em>x</em> ³ - 1 = 0

<em>x</em> ³ - 1³ = 0

Factorize the right side as a difference of cubes:

(<em>x</em> - 1) (<em>x</em> ² + <em>x</em> + 1) = 0

Then

<em>x</em> - 1 = 0   <u>or</u>   <em>x</em> ² + <em>x</em> + 1 = 0

The first equation yields <em>x</em> = 1 as a root.

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

<em>x</em> ² + <em>x</em> + 1/4 = -3/4

(<em>x</em> + 1/2)² = -3/4

<em>x</em> + 1/2 = ±√(-3/4) = ± <em>i</em> √3/2

<em>x</em> = -1/2 ± <em>i</em> √3/2

<em>x</em> = (-1 ± <em>i</em> √3)/2

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

(-1 + <em>i</em> √3)/2 + (-1 - <em>i</em> √3)/2 = (-1 - 1)/2 = -2/2 = -1

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

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