Question

x=3+2√2then find value of (x-1/x)^3​

Answer 1

$\huge\rm\color{PALEVIOLETRED} Solution$

$\sf{x = 3 + 2 \sqrt{2} }$

$\sf{ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } }$

rationalise the denominator.

$\sf{ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } }$

$\sf{ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{ {3}^{2} - {( 2 \sqrt{2})}^{2} } }$

$\sf{ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} }$

$\sf{ \frac{1}{x}} = 3 - 2 \sqrt{2}$

now,

$\sf{x - \frac{1}{x} = 3 + 2 \sqrt{2} - (3 - 2 \sqrt{2}) }$

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

$\sf{x - \frac{1}{x} = 4 \sqrt{2}}$

Hence

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

$\fbox{\sf{({x - \frac{1}{x})}^{3} =128 \sqrt{2} }}$