Math, asked by sushmalvaddi, 3 months ago

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

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Answers

Answered by Anonymous
7

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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