SATHEE: Complex Numbers 5 Question 18

Complex Numbers 5 Question 18

19.

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 $(x + 1)^{n} - x^{n} - 1$

(1980, 3M)

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Solution:

Since, $n$ is not a multiple of 3 , but odd integers and

$x^{3} + x^{2} + x = 0 \Rightarrow x = 0, ω, ω^{2}$

Now, when $x = 0$

$\Rightarrow (x + 1)^{n} - x^{n} - 1 = 1 - 0 - 1 = 0$

$∴ x = 0$ is root of $(x + 1)^{n} - x^{n} - 1$

Again, when $x = ω$

$\Rightarrow (x + 1)^{n} - x^{n} - 1 = (1 + ω)^{n} - ω^{n} - 1 = - ω^{2 n} - ω^{n} - 1 = 0$

[as $n$ is not a multiple of 3 and odd]

Similarly, $x = ω^{2}$ is root of $(x + 1)^{n} - x^{n} - 1$

Hence, $x = 0, ω, ω^{2}$ are the roots of $(x + 1)^{n} - x^{n} - 1$

Thus, $x^{3} + x^{2} + x$ divides $(x + 1)^{n} - x^{n} - 1$ .

Complex Numbers 5 Question 18

19.

Solution:

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Complex Numbers 5 Question 18

19.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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