**Problem:**

If 1, are the nth roots of unity, then find the value of .

**Solution:**

**Learning to think: **

Compare it with what we know from our higher algebra — suppose we have to multiply out:

. We know it is equal to the following:

If we examine the way in which the partial products are formed, we see that

(1) the term is formed by taking the letter x out of each of the factors.

(2) the terms involving are formed by taking the letter x out of any three factors, in every possible way, and one of the letters a, b, c, d out of the remaining factor

(3) the terms involving are formed by taking the letter x out of any two factors, in every possible way, and two of the letters a, b, c, d out of the remaining factors

(4) the terms involving x are formed by taking the letter x out of any one factor, and three out of the letters a, b, c, d out of the remaining factors.

(5) the term independent of x is the product of all the letters a, b, c, d.

**Further hint: **

relate the above to sum of binomial coefficients.

and, you are almost done.

More later,

Nalin Pithwa

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## 2 Comments

Roots of the polynomial (2-x)^n=1 will be 2-w , 2-w^2 …

Hence product of the numbers will be 2^n-1 or 1-2^n depending on the value of n(It will decide the coefficient of x^n.

I agree that the root of is . But, does not seem obviously the root of . Any way, what is the final answer as per your reasoning?