Question

(i) যদি x = 1 + root2 তেন্তে দেখুওৱা যে — (x-1/x)³ = 8​

Answer 1

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$x = 1 + \sqrt{2} \\ \frac{1}{x} = \frac{1}{1 + \sqrt{2} } \\ = \frac{1}{1 + \sqrt{2} } \times \frac{1 - \sqrt{2} }{1 - \sqrt{2} } \\ = \frac{1 - \sqrt{2} }{ {1}^{2} - ( { \sqrt{2} })^{2} } \\ = \frac{1 - \sqrt{2} }{ - 1} \\ = \sqrt{2} - 1 \\ \therefore \: \: \: x - \frac{1}{x} = 1 + \sqrt{2} - \sqrt{2} + 1 \\ = 2 \\ {(x - \frac{1}{x} })^{3} = {2}^{3} = 8$