Question

if x=2-√3 find the value of x3+1/x3

Answer 1

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Answer 2

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$\underline{SOLUTION}$ ➡


$given \: : \: \: x = 2 - \sqrt{3} \\ \\ to \: \: find \: : \: x {}^{3} + \frac{1}{x {}^{3} } =$





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

$x + \frac{1}{x} = (2 - \sqrt{3} ) + (2 + \sqrt{3} ) \\ \\ = > x + \frac{1}{x} = 4 \\ \\ cubing \: \: both \: \: sides \: \: we \: \: get \\ \\ (x + \frac{1}{x } ) {}^{3} = (4) {}^{3} \\ \\ = > (x) {}^{3} + ( \frac{1}{x} ) {}^{3} + 3 \times x \times \frac{1}{x} (x + \frac{1}{x}) = 64 \\ \\ = > x {}^{3} + \frac{1}{x {}^{3} } + 3 \times 4 = 64 \\ \\ = > x {}^{3} + \frac{1}{x {}^{3} } + 12 = 64 \\ \\ = > x {}^{3} + \frac{1}{x {}^{3} } = 64 - 12 \\ \\ = > x {}^{3} + \frac{1}{x {}^{3} } = 52$


$\underline{ANSWER}$ ➡

$x {}^{ 3} + \frac{1}{x {}^{3} } = 52$

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