Question

if x=2-√3, find the value of (x-1\x) cube​

Answer 1

Given :-

x= 2-√3

To find :-

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

A.T.Q :-

$\sf \dfrac{1}{x}=\dfrac{1}{2-\sqrt{3}}\\ \\ \longmapsto \sf \dfrac{1}{x}= \dfrac{1}{2-\sqrt{3}}\times \dfrac{2+\sqrt{3}}{2+\sqrt{3}}\\ \\ \longmapsto\sf \dfrac {1}{x}= \dfrac{2+\sqrt{3}}{(2)^2-(\sqrt{3})^2}\\ \\ \longmapsto\sf \dfrac{1}{x}=\dfrac{2+\sqrt{3}}{4-3}\\ \\\longmapsto\sf \dfrac{1}{x} = \dfrac{2+\sqrt{3}}{1}\\ \\\longmapsto\sf \dfrac{1}{x}= 2+\sqrt{3}$

Now the value of $\sf \bigg(x-\dfrac{1}{x}\bigg)$

We have :-

$\sf x=2-\sqrt{3}\ \ \ \ \ \dfrac{1}{x} =2+\sqrt{3}\\ \\\longmapsto\sf \bigg(x-\dfrac{1}{x}\bigg)= [2-\sqrt{3} -(2+\sqrt{3})]\\ \\\longmapsto\sf \bigg(x-\dfrac{1}{x}\bigg)= (\cancel{2}-\sqrt{3} \cancel{-2}-\sqrt{3})\\ \\ \longmapsto\sf \bigg(x-\dfrac{1}{x}\bigg)= -2\sqrt{3}$

So find the value of $\sf \bigg(x-\dfrac{1}{x}\bigg)^3$

$\longmapsto\sf \bigg(x-\dfrac{1}{x}\bigg)^3= (-2\sqrt{3})^3\\ \\\longmapsto\sf \bigg(x-\dfrac{1}{x}\bigg)^3= ( -2\sqrt{3}\times -2\sqrt{3}\times -2\sqrt{3})\\ \\ \longmapsto\sf \bigg(x-\dfrac{1}{x}\bigg)^3 = (-8\times 3\sqrt{3})\\ \\ \longmapsto\sf \bigg(x-\dfrac{1}{x}\bigg)^3= -24\sqrt{3}$

$\bigstar{\boxed{\sf{\bigg(x-\dfrac{1}{x}\bigg)^3= -24\sqrt{3}}}}$

Answer 2

Answer:

$- 24 \sqrt{3}$

Step-by-step explanation:

$x = 2 - \sqrt{3}$

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

Now we need to rationalise the following,

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

$(a - b)(a + b) = a ^{2} - b ^{2}$

$\frac{1}{x} = \frac{2 + \sqrt{3} }{4 - 3}$

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

Now we need to find the value of x-1/x and cube the value,

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

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

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

Now cubing this value,

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

$(x - \frac{1}{x} ) ^{3} = - 24 \sqrt{3}$