Question

if x=2-underoot3 find x³-1/x³ step by step solution​

Answer 1

Given:

$x = 2 - \sqrt{3}$

To find:

The value of

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

Solution:

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

By using identity

a² - b² = (a+b)(a-b)

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

Now,

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

To find the value of$\: {x}^{3} - \frac{1}{ {x}^{3} }$

we will use the following identity

(x-y)³ = x³ - y³ -3xy(x-y)

=> x³ - y³ = (x-y)³ + 3xy(x-y)

$\small\implies \: {x}^{3} - \frac{1}{ {x}^{3} } = (x - \frac{1}{x} ) {}^{3} + 3 \times \cancel{x} \times \frac{1}{ \cancel{x} }(x - \frac{1}{x} )$

Substituting values from above

We have,

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

So,

$= > {x}^{3} - \frac{1}{ {x}^{3} } = ( - 2 \sqrt{3} ) {}^{3} + 3 \times ( - 2 \sqrt{3} ) \\ \\ = - 24 \sqrt{3} \: - 6 \sqrt{3} \\ \\ = - 30 \sqrt{3}$

Hope it helps you.

Mark as brainliest