Question

if x=2-√3 find the value of x3+1/x3

Answer 1

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So, Thé Answer Is --) 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.

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Answer 2

Hey mate !

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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,

$x + \frac{1}{x} = 2 - \cancel {\sqrt{3}} + 2 + \cancel {\sqrt{3}} \\ \\ x + \frac{1}{x} = 2 + 2 \\ \\ x + \frac{1}{x} = 4$

And,

On cubing both sides.

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


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Thanks for the question!

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