Question

plzz guys help me with this sum..No spam​

Answer 1

Given:-

• x = 1 - √2

To Find:-

• The value of $\sf{\bigg(x-\dfrac{1}{x}\bigg)^3}$

Solution:-

As we are given with the value of x we need to find the value of $\sf{\dfrac{1}{x}}$

Now,

$\sf{\because x = 1 - \sqrt{2}}$,

$\sf{\therefore \dfrac{1}{x} = \dfrac{1}{1 - \sqrt{2}}}$

Now,

We need to rationalize the denominator.

Hence,

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

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

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

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

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

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

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

Now,

We need to find the value of:-

$\sf{\bigg(x - \dfrac{1}{x}\bigg)^3}$

Here we have:-

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

And

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

So we put these values in the equation:-

$\sf{[(1 - \sqrt{2}) - (-\sqrt{2} - 1)]^3}$

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

= $\sf{(1 + 1)^3}$

= $\sf{(2)^3}$

= $\sf{8}$

Hence,

The value of $\sf{\bigg(x - \dfrac{1}{x}\bigg)^3 = 8}$

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