New 55% Off on All Courses - Enroll Now! Click Here

    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 is
    B 1 Correct Answer Incorrect Answer
    C 5 Correct Answer Incorrect Answer
    D 4 Correct Answer Incorrect Answer

    Solution

    (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

    Practice Next

    Relevant for Exams: