Question

if x=2 - √3,then find the value of x^3 - 1/x^3

Answer 1

$answer$

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!