De moivre's theorem & binomial theorem: If x+1/x= 2cosa, prove that x^6-2x^3cos3a+1=0

Math lover799

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De moivre's theorem & binomial theorem: If x+1/x= 2cosa, prove that x^6-2x^3cos3a+1=0

If x+1/x= 2cosa, prove that x^6-2x^3cos3a+1=0 using de moivre's theorem and binomial theorem
 
If x+1/x= 2cosa, prove that x^6-2x^3cos3a+1=0 using de Moivre's Theorem and the Binomial Theorem.
When you reply with the previously-requested information, please also clarify the expressions in the equations.

On the left-hand side of the first equation, you have posted "x+1/x". Do you mean either of the following?

. . . . .\(\displaystyle \mbox{a. }\, x\, +\, \dfrac{1}{x}\)

. . . . .\(\displaystyle \mbox{b. }\, \dfrac{x\, +\, 1}{x}\)

In the middle term of the left-hand side of the second equation, you have included "cos3a". Do you mean either of the following?

. . . . .\(\displaystyle \mbox{c. }\, \cos^3(a)\)

. . . . .\(\displaystyle \mbox{d. }\, \cos(3a)\)

Thank you! ;)
 
I meant x + 1/x and cos3a.. please reply the answer or atleast give a hint. I have tried my level best and I have exams coming up.
 
This is what I have tried..
(X + 1/x) = 2cosa
Cubing both sides and expanding LHS
X^3 +3x + 3/x + 1/(x)^3 = 8cos ^3 (a)
Multiplying throughout by x^3
X^6 + 3x^4 + 3x^2 + 1 = (2cos(a))^3 . x^3
I'm having difficulty after this..
 
To proceed:

My next step would be to use a trigonometric identity for cosine-cubed. Then I'd substitute from the original equation. Rearrangement should then lead to the desired result. ;)
 
(By the way, for some reason, attempting to post to this thread kept crashing the server earlier today. I have no idea what the problem was. My reply was posted elsewhere, and then merged here. I'm hoping the issue is fixed now.)
 
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