Question

if 1,a,a^2,...a^n-1 are nth roots of unity , then the value of (3-a) (3-a^2)....(3-a^n-1) is :​

Answer 1

Given data:

$\mathsf{1,a,a^{2},...,a^{n-1}}$ are $\mathsf{n}$-th roots of unity

To find:

The value of $\mathsf{(3-a)(3-a^{2})...(3-a^{n-1})}$

Step-by-step explanation:

Since $\mathsf{1,a,a^{2},...,a^{n-1}}$ are $\mathsf{n}$-th roots of unity, we can take a variable $\mathsf{x}$ and write the following relation,

$\mathsf{\quad x^{n}-1=(x-1)(x-a)(x-a^{2})...(x-a^{n-1})}$

$\mathsf{\Rightarrow (x-1)(x-a)(x-a^{2})...(x-a^{n-1})=x^{n}-1}$

$\mathsf{\Rightarrow (x-a)(x-a^{2})...(x-a^{n-1})=\frac{x^{n}-1}{x-1}}$ since $\mathsf{x\neq 1}$

Putting $\mathsf{x=3}$ in both sides of the above relation, we get

$\mathsf{\quad (3-a)(3-a^{2})...(3-a^{n-1})=\frac{3^{n}-1}{3-1}}$

$\mathsf{\Rightarrow (3-a)(3-a^{2})...(3-a^{n-1})=\frac{3^{n}-1}{2}}$

$\mathsf{\Rightarrow (3-a)(3-a^{2})...(3-a^{n-1})=\frac{1}{2}(3^{n}-1)}$

Answer:

$\mathsf{(3-a)(3-a^{2})...(3-a^{n-1})=\frac{1}{2}(3^{n}-1)}$