Question

If a₁​,a₂​,a₃​ are the nth roots of unity, then find the value of (w−a₁​)(w−a₂​), where w is a non-real cube root of unity.

a) w+1
b) w−1
c) w²+w+1
d) w²−w+1

Answer 1

The value of (w-a₁)(w-a₂) where w is a non-real cube root of unity is found to be w² - w + 1, using the properties of roots of unity and algebraic manipulation.

Step-by-step explanation:

To find the value of (w-a₁)(w-a₂), where w is a non-real cube root of unity and a₁, a₂, a₃ are the nth roots of unity (specifically, the 3rd roots of unity for this problem), we utilize the properties of roots of unity.

Since w is a non-real cube root of unity, it satisfies the equation w³ = 1.

The cube roots of unity are 1, w, and w², which means that a₁, a₂, and a₃ can be 1, w, and w².

Noting that w ≠ 1 (since w is non-real), we can then assume a₁ = 1, a₂ = w, and a₃ = w².

The product (w - a₁)(w - a₂) is thus (w - 1)(w - w) which simplifies to w(w - 1).

The expanded form of this expression is w² - w.

However, since w is a root of w³ = 1, w² can also be expressed as -w - 1 using the relation w² + w + 1 = 0 (which is derived from w³ - 1 = 0 by factoring out w - 1).

Substituting this into the expression gives us (-w - 1) - w, which simplifies to -2w - 1.

However, since we know that w² + w + 1 = 0, we can substitute -1 for w² + w, yielding the final result w² - w + 1.

Answer 2

The value of (w-a₁)(w-a₂) where w is a non-real cube root of unity is found to be w² - w + 1, using the properties of roots of unity and algebraic manipulation.

Step-by-step explanation:

To find the value of (w-a₁)(w-a₂), where w is a non-real cube root of unity and a₁, a₂, a₃ are the nth roots of unity (specifically, the 3rd roots of unity for this problem), we utilize the properties of roots of unity.

Since w is a non-real cube root of unity, it satisfies the equation w³ = 1.

The cube roots of unity are 1, w, and w², which means that a₁, a₂, and a₃ can be 1, w, and w².

Noting that w ≠ 1 (since w is non-real), we can then assume a₁ = 1, a₂ = w, and a₃ = w².

The product (w - a₁)(w - a₂) is thus (w - 1)(w - w) which simplifies to w(w - 1).

The expanded form of this expression is w² - w.

However, since w is a root of w³ = 1, w² can also be expressed as -w - 1 using the relation w² + w + 1 = 0 (which is derived from w³ - 1 = 0 by factoring out w - 1).

Substituting this into the expression gives us (-w - 1) - w, which simplifies to -2w - 1.

However, since we know that w² + w + 1 = 0, we can substitute -1 for w² + w, yielding the final result w² - w + 1.