Question

if X equal 2 + root 3 then find the value of x cube + 1 by x cube

Answer 1

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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