Question

If (x+1/x)^2=3,then what will be the value of (x)^206+(x)^200+(x)^90+(x)^84+(x)^18+(x)^12+(x)^6+1

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Answer 1

Answer:

(x+1/x)^2=3

or. x^2+ 2 + 1/x^2= 3

or. x^2 - 1 + 1/x^2=0

or. x^4– x^2 + 1 = 0. , Multiplying both sides by (x^2+1).

or. (x^2+1).(x^4-x^2+1)= 0×(x^2+1).

or. (x^6 +1)= 0…………..(1)

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

=x^200.(x^6+1)+x^84.(x^6+1)+x^12.(x^6+1)+1.(x^6+1).

=(x^6 +1).(x^200+x^84+x^12+1). , putting (x^6 +1) =0 from eqn (1).

= 0 × (x^200+x^84+ x^12 +1).

= 0. Answer.

Step-by-step explanation:

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