If 1, 𝜔,𝜔2 are cube roots of unity, show that(2 – 𝜔) (2 – 𝜔2) (2 – 𝜔19) (2 – 𝜔23) = 49.

Sure, let's solve this step by step.

Step 1: We know that the cube roots of unity are 1, 𝜔, 𝜔². These are the roots of the equation x³ = 1.

Step 2: We also know that 𝜔³ = 1, so 𝜔⁴ = 𝜔, 𝜔⁵ = 𝜔², 𝜔⁶ = 1, and so on. This means that we can express 𝜔¹⁹ and 𝜔²³ in terms of 𝜔 and 𝜔².

Step 3: 𝜔¹⁹ = 𝜔⁶3 + 1 = 𝜔¹ = 𝜔 and 𝜔²³ = 𝜔⁶3 + 2 = 𝜔².

Step 4: Now we substitute 𝜔¹⁹ and 𝜔²³ in the given expression with 𝜔 and 𝜔² respectively. So, (2 – 𝜔) (2 – 𝜔²) (2 – 𝜔¹⁹) (2 – 𝜔²³) becomes (2 – 𝜔) (2 – 𝜔²) (2 – 𝜔) (2 – 𝜔²).

Step 5: This simplifies to [(2 – 𝜔)²] [(2 – 𝜔²)²].

Step 6: We know that 1 + 𝜔 + 𝜔² = 0, so 𝜔² = -1 - 𝜔 and 𝜔 = -1 - 𝜔².

Step 7: Substituting these values in, we get [(2 – (-1 - 𝜔²))²] [(2 – (-1 - 𝜔))²] = (3 + 𝜔²)² (3 + 𝜔)².

Step 8: Expanding this, we get (9 + 6𝜔² + 𝜔⁴) (9 + 6𝜔 + 𝜔²).

Step 9: Substituting 𝜔⁴ = 𝜔 and 𝜔² = -1 - 𝜔, we get (9 + 6(-1 - 𝜔) + 𝜔) (9 + 6𝜔 - 1 - 𝜔).

Step 10: Simplifying this, we get (8 - 5𝜔) (8 + 5𝜔).

Step 11: Finally, multiplying this out, we get 64 - 25𝜔².

Step 12: Substituting 𝜔² = -1 - 𝜔, we get 64 - 25(-1 - 𝜔) = 64 + 25 + 25𝜔 = 89 + 25𝜔.

Step 13: But we know that 𝜔 + 1 = -𝜔², so 89 + 25𝜔 = 89 - 25𝜔² = 89 - 25(-1 - 𝜔) = 89 + 25 + 25𝜔 = 114 + 25𝜔.

Step 14: Repeating this process, we get 114 + 25(-1 - 𝜔) = 114 - 25 - 25𝜔 = 89 - 25𝜔.

Step 15: This process will continue indefinitely, so the expression does not simplify to a constant value. Therefore, the given statement is not true. The expression (2 – 𝜔) (2 – 𝜔²) (2 – 𝜔¹⁹) (2 – 𝜔²³) does not equal 49.