Question

For maths answerers!!

Prove that :-

$if \: x = \frac{5 - \sqrt{21} }{2}$

then,

$({x}^{3} + \frac{1}{ {x}^{3} } ) - 5( {x}^{2} + \frac{1}{ {x}^{2} } ) \\ + (x + \frac{1}{x} ) = 0$

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Answer 1

Proof :

Given, $\mathrm{x=\frac{5-\sqrt{21}}{2}}$

Now, $\mathrm{\frac{1}{x}=\frac{2}{5-\sqrt{21}}}$

$\mathrm{=\frac{2(5+\sqrt{21})}{(5-\sqrt{21})(5+\sqrt{21})}}$

$\mathrm{=\frac{2(5+\sqrt{21})}{25-21}}$

$\mathrm{=\frac{2(5+\sqrt{21})}{4}}$

$\mathrm{=\frac{5+\sqrt{21}}{2}}$

Then, $\mathrm{x+\frac{1}{x}}$

$\mathrm{=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}}$

$\mathrm{=\frac{5-\sqrt{21}+5+\sqrt{21}}{2}}$

$\mathrm{=\frac{10}{2}=5}$

• $\mathrm{x^{3}+\frac{1}{x^{3}}}$

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

$\mathrm{=5^{3}-3.5}$

$\mathrm{=125-15=110}$

• $\mathrm{x^{2}+\frac{1}{x^{2}}}$

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

$\mathrm{=5^{2}-2=25-2=23}$

Thus, $\small{\mathrm{(x^{3}+\frac{1}{x^{3}})-5(x^{2}+\frac{1}{x^{2}})+(x+\frac{1}{x})}}$

$\mathrm{=110-5(23)+5}$

$\mathrm{=110-115+5=0}$

$\boxed{\small{\mathrm{(x^{3}+\frac{1}{x^{3}})-5(x^{2}+\frac{1}{x^{2}})+(x+\frac{1}{x})=0}}}$

Hence, proved.

Answer 2

Answer :

Given :

$\tt{x = \frac{5 - \sqrt{21} }{2}}$

We can say that :

$\tt{ \frac{1}{x} = \frac{2}{5 - \sqrt{21}} }$

Rationalizing the denominator,

$\tt{ \frac{1}{x} = \frac{2}{5 - \sqrt{21} } \times \frac{5 + \sqrt{21} }{5 + \sqrt{21}} }$

$\tt{ \frac{1}{x } = \frac{2 \times (5 + \sqrt{21}) }{(5 - \sqrt{21})(5 + \sqrt{21)} }}$

$\tt{\frac{1}{x} = \frac{2 \times (5 + \sqrt{21}) }{25 - 21}}$

$\tt{\frac{1}{x} = \frac{2(5 + \sqrt{21} )}{4}}$

$\tt{ \frac{1}{x} = \frac{5 + \sqrt{21}} {2}}$

Now,

$\tt {x + \frac{1}{x} = \frac{5 - \sqrt{21} }{2} + \frac{5 + \sqrt{21} }{2}}$

$\tt{\therefore x + \frac{1}{x} = 5}$

Squaring both sides ,

$\tt{ ({x} + \frac{1}{x}) {}^{2} = 25}$

$\tt{x {}^{2} + \frac{1}{x {}^{2} } + 2 = 25}$

$\tt{x {}^{2} + \frac{1}{x {}^{2} } = 23}$

Now,

We know that,

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

$= 5 \times (23 - 1)$

$\tt{ = 110}$

Now,

Following next steps ,

$\tt \small{x {}^{3} + \frac{1}{x {}^{3} } - 5 \times (x {}^{2} + \frac{1}{ {x}^{2} }) + (x + \frac{1}{x}) }$

= 110 - (5 × 23) + 5

= 110 - 115 + 5

= 110 - 110

= 0

$\tt{= R. H. S }$

Hence, Proved.