Question

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

Answer 1

Answer:

$- 24 \sqrt{3}$

Step-by-step explanation:

Given:

$x = 2 - \sqrt{3}$

To find :

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

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

Now we need to rationalise the following value to make it easier to solve,

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

In the denominator we can use the identity,

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

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

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

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

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

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

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

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

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

Now cubing this value,

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

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

$- 24 \sqrt{3}$

Answer 2

$= 2 - \sqrt{3} \\ \\ = > \dfrac{1}{x} = 2 + \sqrt{3} \\ \\ = > x - \dfrac{1}{x} = - 2 \sqrt{3}$

Cubing ,

=> - 8 × 3 √ 3

=> -24√3