Question

If x=1-√2,find the value of (x-1/x)^3​

Answer 1

Answer:

$56-8\sqrt{2}$

Step-by-step explanation:

Given:-

$x=1-\sqrt{2}$

To find:-

$(x-\frac{1}{x})^3$

Solution:-

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

On rationalizing $\frac{1}{x}$,

⇒ $\frac{1}{1-\sqrt{2} }$ × $\frac{1+\sqrt{2}}{1+\sqrt{2} }$

⇒  $\frac{1+\sqrt{2}}{(1-\sqrt{2})(1+\sqrt{2}) }$

⇒ $\frac{1+\sqrt{2}}{(1)^2-(\sqrt{2})^2 }$ $($∵ $(a+b)(a-b)=a^2-b^2)$

⇒ $\frac{1+\sqrt{2}}{1-2 }$

⇒ $\frac{1+\sqrt{2}}{-1 }$

⇒ $-(1+\sqrt{2} )$

⇒ $\sqrt{2}-1$

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

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

⇒ $[1-\sqrt{2}-\sqrt{2}+1)]^3$

⇒ $[2-2\sqrt{2} ]^3$

⇒ $(2)^3+(2\sqrt{2} )^3-3$ × $2$ × $2$ × $2\sqrt{2}$ $+3$ × $2$ × $2\sqrt{2}$ × $2\sqrt{2}$

$[$∵ $(a-b)^3=a^3+b^3-3a^2b+3ab^2]$

⇒ $8+16\sqrt{2}-24\sqrt{2}+48$

⇒ $56-8\sqrt{2}$

∴ $(x-\frac{1}{x})^3=56-8\sqrt{2}$