# If x=2+√3 then find x cube +1/x cube what is the answer please tell me

Anoop
12 Points
7 years ago
x = 2 + ∫3  , 1/x = 2-∫3
So, x + 1/x = 4
(x+1/x)3  =  43
x3 + 1/x3 + 3x + 1/x = 64
x3 + 1/x3 + 3 = 64
x3 + 1/x3 =  64 – 3 = 61
praneethvikram
46 Points
7 years ago
given x = 2+ sqrt (3)  =>  1/x will be 2- sqrt(3)

=> x + 1/x = 4
cubing on both sides
=> x^3 + 1/x^3 = 64-3 = 61
Arun
25750 Points
7 years ago
if we find
x+1/x = 4
now (x + 1/x)3 = x3 + 1/x3 + 3* x* 1/x (x + 1/x)
43 = x3 + 1/x3 + 3* 4
so
x3 + 1/x3 = 64 –12 = 52
praneethvikram
46 Points
7 years ago

given x = 2+ sqrt (3)  =>  1/x will be 2- sqrt(3)

=> x + 1/x = 4
cubing on both sides
=> x^3 + 1/x^3 = 64-3(4) = 52

sorry for the mistake done
Kunal friends
13 Points
7 years ago
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]

Mohd Mujtaba
131 Points
6 years ago
(x+1/x)^3=x^3+1/x^3+3x+3/x . Given x=2+root3. So put value in above we get answer 52. Hope you understand thank ☺☺☺☺☺
yathartha gupta
71 Points
6 years ago
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 =16-12x+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=64-12x3+1/x3=52 [Ans]
C.Vishaal
39 Points
6 years ago
Alright 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

gg
11 Points
6 years ago


x = 2 + √3
1/x = - 23
 . 

( x – 1/x)^3= -24√3

ediots
so remember my advise use the brains not the mouths


C.Vishaal
39 Points
6 years ago
1/x cannot be -2 + root 3, you have to rationalise the Denominator So your answer which is I guess -24/3 is wrong.
Nikhil Chouti
27 Points
5 years ago
x=2+√3
1/x=1/2+√3 which upon rationalising we get 2-√3

x^3+1/x^3=2+√3 + 2-√3
=4

Nikhil Chouti
27 Points
5 years ago
Oh sorry the answer would be 52 since we r cubing
We should use (a+b)^3 formula n (a-b)^3 formula to get solution
Ajeet Tiwari
4 years ago
hello students

since x=1/x
if we find
x+1/x = 4
now (x + 1/x)3= x3+ 1/x3+ 3* x* 1/x (x + 1/x)
43=x3+ 1/x3+ 3* 4
so
x3+ 1/x3= 64 –12 = 52

Hope it helps
Thankyou and Regards