Question

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it is given that n is an odd integer greater than 3 but n is not a multiple of 3. prove that x cube + x square + x is a factor of (x + 1)^n - x^n-1.​

Answer 1

Answer:

from remainder theorem, we know that if x^3 + x^2 + x is a factor of (x+1)^n - x^n - 1, then zeros of x^3+x^2+x must satisfy (x+1)^n - x^n - 1. now we will find zeros of the former and will see if the zeros are also the zeros of the latter

x^3+x^2+x= 0

x(1+x+x^2)= 0

hence either x = 0 or

1+x+x^2 = 0

from cube root of unity we know that solutions of 1+x+x^2 is in imaginary numbers and are denoted by w and w^2 (omega)

hence solution of 1+x+x^2 = 0, w, w^2

now let us check whether these are also the zeros of given polynomial

put x = 0

(0+1)^n - 0^n - 1

1-1 = 0

hence x=0 is a solution of the polynomial

put x = w

(w+1)^n - w^n - 1

(-w^2)^n - w^n - 1 ( 1+w = -w^2)

-(w^n)^2 - w^n - 1 ( since n is odd)

now since n is not multiple of 3, w^n will not be equal to 1. for any other odd integer,the expression will reduce to -(1+w+w^2) which is equal to 0.

hence, x=w is also a solution

put x = w^2

(w^2+1) - (w^2)^n - 1

(-w)^n - (w^n)^2 - 1. (1+w^2 = w)

- w^n - (w^2)^n - 1 ( since n is odd)

this is again the same equation which we got when we put x=w. hence, this is also equal to zero.

so, zeros of x^3 + x^2 + x are also the zeros of (x+1)^n - x^n - 1.

hence x^3+x^2+x is a factor of (x+1)^n - x^n - 1.

Answer 2

Answer:

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