Question

If x= 2+√3, find the value of x³+1/x³

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

HOPE THIS WILL HELP YOU...

Answer 2

Hi there !

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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

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


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Thanks for the question !

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