SOLUTION: if (x+1/x)^2 = 3 then find the value of x^206 + x^200 + x^90 + x^84 + x^18 + x^12 + x^6 + 1

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: if (x+1/x)^2 = 3 then find the value of x^206 + x^200 + x^90 + x^84 + x^18 + x^12 + x^6 + 1       Log On


   

Question 1036823: if (x+1/x)^2 = 3 then find the value of x^206 + x^200 + x^90 + x^84 + x^18 + x^12 + x^6 + 1
Answer by ikleyn(52781) About Me  (Show Source):
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if (x+1/x)^2 = 3 then find the value of x^206 + x^200 + x^90 + x^84 + x^18 + x^12 + x^6 + 1
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Answer. If %28x%2B1%2Fx%29%5E2 = 3 then the value of x%5E206+%2B+x%5E200+%2B+x%5E90+%2B+x%5E84+%2B+x%5E18+%2B+x%5E12+%2B+x%5E6+%2B+1 is zero.

Solution 
If  %28x%2B1%2Fx%29%5E2 = 3  then

x%5E2 + 2 + 1%2Fx%5E2 = 3,

x%5E2 + 1%2Fx%5E2 = 1,

x%5E2+-1+%2B+1%2Fx%5E2 = 0,   ( <--- Introduce new variable u = x%5E2 )  ---> 

u+-+1+%2B+1%2Fu = 0,   ( <---multiply both sides by u)  ---> 

u%5E2+-u+%2B+1 = 0,    ( <--- solve by using the quadratic formula )  ---> 

u%5B1%2C2%5D = 1%2F2+%2B-+i%2A%28sqrt%283%29%2F2%29 = cos%28pi%2F3%29+%2B-+i%2Asin%28pi%2F3%29%29,   --->

x%5B1%2C2%5D = cos%28pi%2F6%29+%2B-+i%2Asin%28pi%2F6%29.   ( since x = sqrt%28u%29 )

In other words, if  %28x%2B1%2Fx%29%5E2 = 3  then x is the primitive root of the degree 12 of 1.

Then

x%5E90 = -x%5E84,  so  x%5E90+%2B+x%5E84 = 0;

x%5E18 = -x%5E12,  so  x%5E18+%2B+x%5E12 = 0;

x%5E6 = -1,  so  x%5E6+%2B+1 = 0,  and also

x%5E206 = -x%5E200,  so  x%5E206+%2B+x%5E200 = 0.

Hence,  the value of x%5E206+%2B+x%5E200+%2B+x%5E90+%2B+x%5E84+%2B+x%5E18+%2B+x%5E12+%2B+x%5E6+%2B+1 is zero.
Solved.