Question

If x=2-root under3, find the value lf (x+1/x)cube+(x+1/x)square+(x+1/x)-100.

Answer 1

Given,

x = 2 - √3

=> 1/x = 1/(2 - √3)

Rationalising it, we get

1/x = (2+√3) ÷ (2 -√3)(2 + √3)

=> 1/x = 2+ √3/4 - 3

=> 1/x = 2 + √3/1

=> 1/x = 2 + √3


Now we have to find,

( x + 1/x)³ + (x + 1/x)² + (x + 1/x) - 100

Substituting the value of x and 1/x we get,

(2 - √3 + 2 +√3)³ + (2 - √3 + 2 + √3)² + (2 - √3 + 2 + √3) - 100

=> (4)³ + (4)² + (4) - 100

=> 64 + 16 + 4 - 100

=> 84 - 100

=> -16


Hence, your answer is -16


Hope it helps dear friend ☺️