Question

If x = √3-√2 find the value of x³ + 1/ x³ - 3(x²+ 1/ x²) + x + 1/x​

Answer 1

Answer:GIVEN :

x=2+√3  

1/x = 1/ 2+√3

x+1/x =2+√3+1/(2+√3)

x+1/x  =[(2+√3)(2+√3)+1] /2+√3

 [by taking  LCM ]

x+1/x  =[(2+√3)² +1] /2+√3

x+1/x = (2² + √3² + 2×2 ×√3 )+1 / (2+√3)

[ (a+b)² = a² + b² + 2ab ]

x+1/x  = 4+ 3+ 4×√3 +1 /(2+√3)

x+1/x  = 7+1 + 4√3  

x+1/x = 8+4√3/ 2+√3

x+1/x =[8+4√3/(2 +√3)×[2-√3 / 2-√3]    

[by rationalising the denominator]

       =[8+4√3][2-√3] / 2²- √3

[ (a+b)(a - b) = a² - b² ]

       =16 + 8√3 - 8√3 - 4× 3 / 4 - 3

              =16 -12/1 = 4

x+1/x = 4…………… (1)

[x+1/x]³ = 4³    [On cubing both sides]

x³+1/x³+3x×1/x[x+1/x] = 64    

[using the formula  (x+y)³ = x³+ y³ + 3xy(x+y)]

x³ +1/x³+3[x+1/x] = 64    

x³ +1/x³+3×4 = 64     [from eq 1)

x³ +1/x³+ 12 = 64

x³ +1/x³ = 64 -12 = 52

x³ +1/x³ = 52

Hence, the value of x³ +1/x³ = 52

HOPE THIS WILL HELP YOU...

Step-by-step explanation: pls mark me brainliest

Answer 2

Answer: lol i dont known i will copy from previous ans

Step-by-step explanation:GIVEN :

x=2+√3  

1/x = 1/ 2+√3

x+1/x =2+√3+1/(2+√3)

x+1/x  =[(2+√3)(2+√3)+1] /2+√3

 [by taking  LCM ]

x+1/x  =[(2+√3)² +1] /2+√3

x+1/x = (2² + √3² + 2×2 ×√3 )+1 / (2+√3)

[ (a+b)² = a² + b² + 2ab ]

x+1/x  = 4+ 3+ 4×√3 +1 /(2+√3)

x+1/x  = 7+1 + 4√3  

x+1/x = 8+4√3/ 2+√3

x+1/x =[8+4√3/(2 +√3)×[2-√3 / 2-√3]    

[by rationalising the denominator]

       =[8+4√3][2-√3] / 2²- √3

[ (a+b)(a - b) = a² - b² ]

       =16 + 8√3 - 8√3 - 4× 3 / 4 - 3

              =16 -12/1 = 4

x+1/x = 4…………… (1)

[x+1/x]³ = 4³    [On cubing both sides]

x³+1/x³+3x×1/x[x+1/x] = 64    

[using the formula  (x+y)³ = x³+ y³ + 3xy(x+y)]

x³ +1/x³+3[x+1/x] = 64    

x³ +1/x³+3×4 = 64     [from eq 1)

x³ +1/x³+ 12 = 64

x³ +1/x³ = 64 -12 = 52

x³ +1/x³ = 52

Hence, the value of x³ +1/x³ = 52

HOPE THIS WILL HELP YOU...