Question

if x=1-root2,then find the value of (x-1/x) whole cube​

Answer 1

Answer :-

Given

x = 1 - √2

Value to find

(x - 1/x)³

First we will rationalise the denominator of 1/x

$= \dfrac{1}{1-\sqrt{2}}$

$= \dfrac{1}{1-\sqrt{2}} \times \dfrac{1+ \sqrt{2}}{1+\sqrt{2}}$

$= \dfrac{1 + \sqrt{2}}{1^2 - \sqrt{2}^2}$

$=\dfrac{ 1 + \sqrt{2}}{1 - 2}$

$= \dfrac{ (1+\sqrt{2})}{-1}$

$= \dfrac{ -(1+\sqrt{2})}{1}$

Now we have value of 1/x = -(1+ √2)

So ( x - 1/x)³

= ( 1 - √2 - (-(1 + √2)))³

= ( 1 - √2 + 1 + √2)³

= (2)³

= 8