Question

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 1

( x+ 1/x)^2 =3

(x+1/x)=√3

cubing both side

(x)^3+ (1/x)^3 +3(x) +3(1/x) =3√3

(x)^3 +(1/x )^3+ 3(x+ 1/x) = 3√3

(x)^3 +(1/x)^3= 3√3 -3√3

(x)^3 +(1/x)^3= 0

Now

x^206 +x^200 +x^90 + x^84 +x^18 +x^12 x^6 + 1= ?

x^203(x^3 +1/x ^3) + x^87(x^3 +1/x ^3)+ x^15 (x^3 + 1/x ^3) + x^3(x^3+ 1/x ^3)

From the given equation we know that( x^3 + 1/x ^3)= 0

So putting the value gives 0 as an answer.

Hope this helps