Question

If x+1/x=root 3 find the value of x cube + 1/x cube

Answer 1

Hi there!

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Given :

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

To find :

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

Solution:

$x + \frac{1}{x} = \sqrt{3} \\ \\ \bold{on \: cubing \: both \: sides.} \\ \\ (x + \frac{1}{x} ) {}^{3} = ( \sqrt{3} ){}^{3} \\ \\ x{}^{3} + \frac{1}{x{}^{3}} + 3(x + \frac{1}{x} ) = 3 \sqrt{3} \\ \\ x{}^{3} + \frac{1}{x{}^{3}} + 3 \times \sqrt{3} = 3 \sqrt{3} \\ \\ x{}^{3} + \frac{1}{x{}^{3}} = 3 \sqrt{3} - 3 \sqrt{3} \\ \\ x{}^{3} + \frac{1}{x{}^{3}} = 0$


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Thanks for the question!

Answer 2

Given: x + 1/x = √3

To find: The value of x³ + 1/x³.

Solution:

Since the value of (x + 1/x) is given in the question, the value of (x² + 1/x²) can be calculated by squaring the term (x + 1/x) and then, rearranging the values in it.

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

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

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

            $= 2$

So, the value of (x² + 1/x²) is found to be 2. Now, the value of (x³ + 1/x³) can be calculated by multiplying the term (x + 1/x) with (x² + 1/x²) and rearranging the terms.

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

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

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

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

             $= 2\sqrt{3} - \sqrt{3}$

             $= \sqrt{3}$

Therefore, the value of x³ + 1/x³ is √3.