Question

if x= √3-√2 then find the value of x cube +1/x cube​

Answer 1

Given:

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

To find:

${x}^{3} + ( \frac{1}{x } ) {}^{3}$ =?

Solution:

$x = \sqrt{3} - \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{ \sqrt{3} - \sqrt{2} } \\ \\ rationalising \: the \: denominator$

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

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

$= \sqrt{3} + \sqrt{2}$

Now,

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

We know that,

[ (a+b)³= a³ + b³ + 3ab (a+b) ]

Using this identity

we get

a³ + b³ = (a+b)³ - (3ab (a+b) )

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

Substituting values

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

$= 24 \sqrt{3} - 6 \sqrt{3} \\ = 18 \sqrt{3}$

Hence, the value is 18√3.

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