Question

if X equal to 2 + root 3 then find the value of x cube + 1 by x cube

Answer 1

if X equal to 2 + root 3 then find the value of x cube + 1 by x cube

x = 2 + ∫3  , 1/x = 2-∫3  

So, x + 1/x = 4

(x+1/x)3  =  43 

x3 + 1/x3 + 3x + 1/x = 64

or

x=2+∫3   [given]

x+1/x =2+∫3+1/2+∫3

         =[2+∫3][2+∫3]+1/2+∫3   [taking LCM ]

         =[2+∫3]2+1/2+∫3

         =4+3+[4×∫3]+1/2+∫3

         =8+4∫3/2+∫3

         =[8+4∫3/2+∫3]×[2-∫3/2-∫3]    [by rationalising the denominator]

         =[8+4∫3][2-∫3]/[2]2-[∫3]2

         =16-8∫3+8∫3-4[∫3]2

            =16-12

x+1/x=4

[x+1/x]3=43     [cubing both side]

x3+1/x3+3×x×1/x[x+1/x]=64     [using {x+y}3]

x3+1/x3+3×4=64     [substituting the value of x+1/x]\

x3+1/x3=64-12

x3+1/x3=52 [Ans]

or

since x = 2 + root 3, upon rationalistaion, we get 1/x as 2 – root 3.

Alright, do { x + 1/x }3 = x3 + 1/x3 + 3 ( x multiplied by 1/x } …......{1}

So that will give you (2 + root 3 + 2 – root 3)3 = 43 = 64

When 3 in {1} is taken to the other side, its sign will become negative ( Due to transposition )

That x3 + 1/x3 = 64 – 3 = 61

Hope this answer helps you.