Algebra and Applications Test 1

Question : 16 of 40

Marks: +1, -0

(x+1/x)^2=3,

then the value of

x^ 206 + x ^200 + x ^90

+ x ^84 + x ^18

+ x ^12 + 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^ 6 + 1 = 0

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) + (x ^6 + 1) = 0

Go to Question:

Prev Question

Next Question