It is given that n is an odd integer greater than 3, but n is not a multiple of 3. Prove that x3+x2+x is a factor of (x+1)n- xn- 1 .

Question

It is given that n is an odd integer greater than 3, but n is not a multiple of 3. Prove that x3+x2+x is a factor of (x+1)n- xn- 1 . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Since, n is not a multiple of 3, but odd integers and x3+x2+x=0 ⇒ x=0, ω, ω2 Now, when x = 0 ⇒ ( x + 1)n - x n - 1 = 1 - 0 - 1 = 0 ∴ x = 0 is root of (x + 1)n- xn- 1 Again, when x = ω ⇒ (x+1)n-xn-1=(1+ω)n-ωn -1 =-ω2n-ωn-1=0 [as n is not a multiple of 3 and odd] Similarly, x = ω2 is root of (x + 1)n - xn - 1 Hence, x = 0,ω, ω2 are the roots of (x + 1)n - xn - 1 Thus, x3+x2+ x divides (x + 1)n - xn - 1 .

Q. It is given that n is an odd integer greater than 3, but n is not a multiple of 3. Prove that x3+x2+x is a factor of (x+1)n−xn−1 .

Q. It is given that $n$ is an odd integer greater than $3$ , but $n$ is not a multiple of $3$ . Prove that $x^{3} + x^{2} + x$ is a factor of $(x + 1)^{n} - x^{n} - 1$ .