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If ω is a cube root of unity, then the linear factors of x3+y3 in complex numbers are
- a)(x+y)(x+yω)(x+yω2)
- b)(x+y)(x−yω)(x−yω2)
- c)(x+y)(x+yω)(x−yω2)
- d)(x−y)(x+yω)(x+yω2)
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Ifωis a cube root of unity, then the linear factors ofx3+y3 in c...
Understanding Cube Roots of Unity
The cube roots of unity are the solutions to the equation x^3 = 1. These solutions are:
- 1
- ω (omega) = e^(2πi/3)
- ω^2 = e^(4πi/3)
Here, ω and ω^2 are complex numbers representing the non-real cube roots of unity.
Factoring x^3 + y^3
The expression x^3 + y^3 can be factored using the identity:
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
This factorization is true for any complex numbers x and y.
Finding the Linear Factors
To express x^3 + y^3 in terms of linear factors, we can consider:
- The roots of unity (ω and ω^2)
- The property that ω^3 = 1 and 1 + ω + ω^2 = 0
We can rewrite y^3 as (y)(y)(y), and substitute in the roots of unity:
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
By setting y = yω and y = yω^2, we find the linear factors:
- (x + y)
- (x + yω)
- (x + yω^2)
Thus, the complete factorization of x^3 + y^3 in terms of linear factors is:
(x + y)(x + yω)(x + yω^2)
Conclusion
The correct answer is option 'A', which presents the linear factors of x^3 + y^3 as (x + y)(x + yω)(x + yω^2). This factorization captures all roots and adheres to the properties of complex numbers and cube roots of unity.
The cube roots of unity are the solutions to the equation x^3 = 1. These solutions are:
- 1
- ω (omega) = e^(2πi/3)
- ω^2 = e^(4πi/3)
Here, ω and ω^2 are complex numbers representing the non-real cube roots of unity.
Factoring x^3 + y^3
The expression x^3 + y^3 can be factored using the identity:
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
This factorization is true for any complex numbers x and y.
Finding the Linear Factors
To express x^3 + y^3 in terms of linear factors, we can consider:
- The roots of unity (ω and ω^2)
- The property that ω^3 = 1 and 1 + ω + ω^2 = 0
We can rewrite y^3 as (y)(y)(y), and substitute in the roots of unity:
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
By setting y = yω and y = yω^2, we find the linear factors:
- (x + y)
- (x + yω)
- (x + yω^2)
Thus, the complete factorization of x^3 + y^3 in terms of linear factors is:
(x + y)(x + yω)(x + yω^2)
Conclusion
The correct answer is option 'A', which presents the linear factors of x^3 + y^3 as (x + y)(x + yω)(x + yω^2). This factorization captures all roots and adheres to the properties of complex numbers and cube roots of unity.
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Ifωis a cube root of unity, then the linear factors ofx3+y3 in complex numbers area)(x+y)(x+yω)(x+yω2)b)(x+y)(x−yω)(x−yω2)c)(x+y)(x+yω)(x−yω2)d)(x−y)(x+yω)(x+yω2)Correct answer is option 'A'. Can you explain this answer?
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