Question
If (x+1/x)^2 = 3, then the value of x^206 + x^200 +
x^90 + x^84 + x^42 + x^36 + x^12 + x^6 + 5 isSolution
(x+1/x)^2= 3 → x+1/x = √3 By cubing both sides, (x+1/x)^3= (√3)^3 x^3+1/x^3 + 3 × x × 1/x (x+1/x) = 3√3 x^3+1/x^3 + 3 × √3× = 3√3 x^3+1/x^3 = 3√3 - 3√3 x^3+1/x^3 = 0 (x^6+1)/x^3 = 0 → x^6+1 = 0 x^206 + x^200 + x^90 + x^84 + x^42 + x^36 + x^12 + x^6 + 5 x^200 (x^6 +1) + x^84 (x^6 +1) + x^36 (x^6 +1) + x^6 (x^6 +1) + 5 By putting the value (x^6 +1=0 ) x^200 × 0 + x^84 × 0 + x^36 × 0 + x^6 × 0 + 5 = 5
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