Question

if x =2+√3 find the value of x^3+1/x^3
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Answer 1

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

Answer 2

Given :

x = 2  + \sqrt{3}  

To find ;

x {}^{3}  +  \frac{1}{x {}^{3} }  

Solution :

x = 2 +  \sqrt{3}  \\  \\  \frac{1}{x}  =  \frac{1}{2 +  \sqrt{3} }  \times  \frac{2  -  \sqrt{3} }{2 -  \sqrt{3} }  \\  \\  \frac{1}{x}  =  \frac{2 -  \sqrt{3} }{(2) {}^{2} -  (\sqrt{3} ) {}^{2}  }  \\  \\  \frac{1}{x}  =  \frac{2 -  \sqrt{3} }{4 - 3}  \\  \\  \frac{1}{x}  = 2 -  \sqrt{3}  

Now,

x +  \frac{1}{x} \\  \\  \implies2 +   \cancel{\sqrt{3}}+ 2 -  \cancel {\sqrt{3}} \\  \\  \implies2 + 2 \\  \\   \implies4

So, on cubing both sides, we have

(x +  \frac{1}{x} ) {}^{3}  = (4){}^{3} \\  \\ x{}^{3} +  \frac{1}{x{}^{3}}  + 3(x +  \frac{1}{x} ) = 64 \\  \\ x{}^{3} +  \frac{1}{x{}^{3}}  +3 \times 4 = 64 \\  \\ x{}^{3} +  \frac{1}{x{}^{3}}  +12 = 64 \\  \\ x{}^{3} +  \frac{1}{x{}^{3}}  = 64 - 12 \\  \\ x{}^{3} +  \frac{1}{x{}^{3}}   = 64 - 12 \\  \\ \boxed{ \bold{ x{}^{3} +  \frac{1}{x{}^{3}}   = 52}}