Question

if x=(5-√21)/2 , prove that (x^3 + 1/x^3) -5(x^2 + 1/x^2) + (x + 1/x) = 0

Answer 1

x=(5-√21)/2
1/x=2/(5-√21)=2(5+√21)/4=(5+√21)/2

now,
x+1/x=5
x^2+1/x^2=5^2-2=23

x^3+1/x^3=5^3-3*5=110

putting values

110-5*23+5=0

Answer 2

Answer:

⇒110 - 110 = 0 ( proved )

Step-by-step explanation:

Given ;

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

To prove ;

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

Solution ;

Hence ,

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

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

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

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

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

Hence ,

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

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

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

Hence ,

$= > (x + \frac{1}{x} )^{2} = 5^{2} = 25$

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

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

$= > (x^{2} + \frac{1}{x^{2} } ) = 25 - 2 = 23$

And , x + 1/x = 5

$= > (x + \frac{1}{x} )^{3} = (5) ^{3} = 125$

$= > x^{3} + \frac{1}{x^{3} } + 3 \times x \times \frac{1}{x} ( x + \frac{1}{x}) = 125$

$= > (x^{3} + \frac{1}{x^{3} }) = 125 - 15 = 110$

Hence ,

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

⇒110 - 5 × 23 + 5

⇒110 - 115 + 5

⇒110 - 110 = 0