If (x+1/x)^2=3 then the value of (x^ 72 + x^ 66 + x^ 54 + x^ 36 + x ^24 + x ^6 + 1) is:

Algebra and Applications Test 2

Question : 10 of 40

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(x+1/x)^2=3

then the value of

(x^ 72 + x^ 66 + x^ 54 + x^ 36 + x ^24 + x ^6 + 1)

is:

Solution:

(x+1/x)^2=3

x+1/x =√3

Cubing both sides, we get:

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

x^6+1=0

∴ x^ 72 + x^ 66 + x^ 54 + x^ 36 + x ^24 + x ^6 + 1

= (x^ 6 )^ 12 + (x^ 6 )^ 11 + (x^ 6 )^ 9 + (x^ 6 )^ 6 + (x^ 6 )^ 4 + x^6 + 1

= (-1)^12 + (-1)^11 + (-1)^9 + (-1)^6 + (-1)^4 + (-1) + 1

= 1 - 1 - 1 + 1 + 1 + 1 + 1

= 4 - 3 = 1

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