if p−5 = 2(√6) then prove that p(√p)−(1/(p(√p))) = 22(√2)


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Question Number 159842 by abdullah_ff last updated on 21/Nov/21

$if p - 5 = 2 \sqrt{6} then prove that$ $p \sqrt{p} - \frac{1}{p \sqrt{p}} = 22 \sqrt{2}$
Commented by tounghoungko last updated on 21/Nov/21

$f a l s e$
Commented by abdullah_ff last updated on 21/Nov/21

$w h y ?$
Commented by mr W last updated on 21/Nov/21

$y o u h a v e c h a n g e d t h e q u e s t i o n$ $f r o m p - 5 = 2 \sqrt{3} t o p - 5 = 2 \sqrt{6} .$ $w i t h p - 5 = 2 \sqrt{6} i t i s t h e n t r u e .$
Commented by abdullah_ff last updated on 22/Nov/21

$that was a typing mistake . sorry sir .$
	Answered by Rasheed.Sindhi last updated on 21/Nov/21

	$p - 5 = 2 \sqrt{6} \Rightarrow p = 5 + 2 \sqrt{6}$ $\sqrt{p} = \sqrt{3} + \sqrt{2}$ $▸ p \sqrt{p} - \frac{1}{p \sqrt{p}} = {(\sqrt{p})}^{3} - \frac{1}{{(\sqrt{p})}^{3}}$ $= {(\sqrt{3} + \sqrt{2})}^{3} - \frac{1}{{(\sqrt{3} + \sqrt{2})}^{3}}$ $= {(^{★} \sqrt{3} + \sqrt{2} - \frac{1}{\sqrt{3} + \sqrt{2}})}^{3} + 3 (\sqrt{3} + \sqrt{2} - \frac{1}{\sqrt{3} + \sqrt{2}})$ $^{★} \sqrt{3} + \sqrt{2} - \frac{1}{\sqrt{3} + \sqrt{2}}$ $= \sqrt{3} + \sqrt{2} - \frac{1}{\sqrt{3} + \sqrt{2}} . \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$ $= \sqrt{3} + \sqrt{2} - \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = 2 \sqrt{2}$ $= {(2 \sqrt{2})}^{3} + 3 (2 \sqrt{3})$ $= 8.2 . \sqrt{2} + 6 \sqrt{3} = 22 \sqrt{2}$
	Commented by abdullah_ff last updated on 22/Nov/21

	$thank you sir$
	Answered by Rasheed.Sindhi last updated on 22/Nov/21
	$determinant (((Different Way...))) (√p) =(√3) +(√2) ⇒(1/( (√p) ))=(1/( (√3) +(√2))).(((√3) −(√2))/( (√3) −(√2)))=(√3) −(√2) p(√p)−(1/(p(√p)))=((√p) )^3 −((1/( (√p))))^3 =((√3) +(√2))^3 −((√3) −(√2))^3 =((√3) +(√2) −(√3) +2(√2) ) ×{((√3) +(√2))^2 +((√3) +(√2))((√3) −(√2))+((√3) −(√2))^2 } =2(√2) {(3+2+2(√6) )+(3−2)+(3+2−2(√6))} =2(√2) (5+2(√6) +1+5−2(√6)) =2(√2) (11)=22(√2)$
	$\begin{matrix} D i f f e r e n t W a y . . . \end{matrix}$ $\sqrt{p} = \sqrt{3} + \sqrt{2}$ $\Rightarrow \frac{1}{\sqrt{p}} = \frac{1}{\sqrt{3} + \sqrt{2}} . \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \sqrt{3} - \sqrt{2}$ $p \sqrt{p} - \frac{1}{p \sqrt{p}} = {(\sqrt{p})}^{3} - {(\frac{1}{\sqrt{p}})}^{3}$ $= {(\sqrt{3} + \sqrt{2})}^{3} - {(\sqrt{3} - \sqrt{2})}^{3}$ $= (\sqrt{3} + \sqrt{2} - \sqrt{3} + 2 \sqrt{2})$ $\times {{(\sqrt{3} + \sqrt{2})}^{2} + (\sqrt{3} + \sqrt{2}) (\sqrt{3} - \sqrt{2}) + {(\sqrt{3} - \sqrt{2})}^{2}}$ $= 2 \sqrt{2} {(3 + 2 + 2 \sqrt{6}) + (3 - 2) + (3 + 2 - 2 \sqrt{6})}$ $= 2 \sqrt{2} (5 + 2 \sqrt{6} + 1 + 5 - 2 \sqrt{6})$ $= 2 \sqrt{2} (11) = 22 \sqrt{2}$
	Commented by abdullah_ff last updated on 22/Nov/21

	$you are great sir . .$
	Commented by Rasheed.Sindhi last updated on 22/Nov/21

	$abdullah sir, where are you from ?$
	Commented by abdullah_ff last updated on 22/Nov/21

	$Bangladesh . And you sir ?$
	Commented by Rasheed.Sindhi last updated on 22/Nov/21

	$From Sindh Pakistan .$
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