if-p-5-2-6-then-prove-that-p-p-1-p-p-22-2-

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

$$\mathrm{if}\:{p}−\mathrm{5}\:=\:\mathrm{2}\sqrt{\mathrm{6}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${p}\sqrt{{p}}−\frac{\mathrm{1}}{{p}\sqrt{{p}}}\:=\:\mathrm{22}\sqrt{\mathrm{2}} \\ $$

Commented by tounghoungko last updated on 21/Nov/21

$${false} \\ $$

Commented by abdullah_ff last updated on 21/Nov/21

$${why}? \\ $$

Commented by mr W last updated on 21/Nov/21

you have changed the question from p−5=2(√3) to p−5=2(√6). with p−5=2(√6) it is then true.

$${you}\:{have}\:{changed}\:{the}\:{question}\: \\ $$$${from}\:{p}−\mathrm{5}=\mathrm{2}\sqrt{\mathrm{3}}\:{to}\:{p}−\mathrm{5}=\mathrm{2}\sqrt{\mathrm{6}}. \\ $$$${with}\:{p}−\mathrm{5}=\mathrm{2}\sqrt{\mathrm{6}}\:{it}\:{is}\:{then}\:{true}. \\ $$

Commented by abdullah_ff last updated on 22/Nov/21

$$\mathrm{that}\:\mathrm{was}\:\mathrm{a}\:\mathrm{typing}\:\mathrm{mistake}.\:\mathrm{sorry}\:\mathrm{sir}. \\ $$

Answered by Rasheed.Sindhi last updated on 21/Nov/21

p−5=2(√6) ⇒p=5+2(√6) (√p)=(√3) +(√2) ▶p(√p)−(1/(p(√p)))=((√p) )^3 −(1/(((√p) )^3 )) =((√3) +(√2) )^3 −(1/( ((√3) +(√2) )^3 )) =(^★ (√3) +(√2) −(1/( (√3) +(√2) )))^3 +3((√3) +(√2) −(1/( (√3) +(√2) ))) ^★ (√3) +(√2) −(1/( (√3) +(√2) )) =(√3) +(√2) −(1/( (√3) +(√2) )).(((√3) −(√2))/( (√3) −(√2))) =(√3) +(√2) −(((√3) −(√2))/(3−2))=2(√2) =(2(√2) )^3 +3(2(√3) ) =8.2.(√2) +6(√3)=22(√2)

$${p}−\mathrm{5}=\mathrm{2}\sqrt{\mathrm{6}}\:\Rightarrow{p}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\: \\ $$$$\sqrt{{p}}=\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\: \\ $$$$\blacktriangleright{p}\sqrt{{p}}−\frac{\mathrm{1}}{{p}\sqrt{{p}}}=\left(\sqrt{{p}}\:\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\left(\sqrt{{p}}\:\right)^{\mathrm{3}} } \\ $$$$=\left(\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\:\left(\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:\right)^{\mathrm{3}} } \\ $$$$=\left(^{\bigstar} \sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:}\right)^{\mathrm{3}} +\mathrm{3}\left(\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:^{\bigstar} \sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:}.\frac{\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:−\frac{\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}}{\mathrm{3}−\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$$=\left(\mathrm{2}\sqrt{\mathrm{2}}\:\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{2}\sqrt{\mathrm{3}}\:\right) \\ $$$$=\mathrm{8}.\mathrm{2}.\sqrt{\mathrm{2}}\:+\mathrm{6}\sqrt{\mathrm{3}}=\mathrm{22}\sqrt{\mathrm{2}} \\ $$

Commented by abdullah_ff last updated on 22/Nov/21

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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)

$$\begin{array}{|c|}{\mathcal{D}{ifferent}\:\mathcal{W}{ay}…}\\\hline\end{array} \\ $$$$\sqrt{{p}}\:=\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\: \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{p}}\:}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}}.\frac{\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}}=\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}} \\ $$$${p}\sqrt{{p}}−\frac{\mathrm{1}}{{p}\sqrt{{p}}}=\left(\sqrt{{p}}\:\right)^{\mathrm{3}} −\left(\frac{\mathrm{1}}{\:\sqrt{{p}}}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:=\left(\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\right)^{\mathrm{3}} −\left(\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$=\left(\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}\:+\mathrm{2}\sqrt{\mathrm{2}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left\{\left(\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}\right)+\left(\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \right\} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\:\left\{\left(\mathrm{3}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{6}}\:\right)+\left(\mathrm{3}−\mathrm{2}\right)+\left(\mathrm{3}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{6}}\right)\right\} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\:\left(\mathrm{5}+\cancel{\mathrm{2}\sqrt{\mathrm{6}}}\:+\mathrm{1}+\mathrm{5}−\cancel{\mathrm{2}\sqrt{\mathrm{6}}}\right) \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\:\left(\mathrm{11}\right)=\mathrm{22}\sqrt{\mathrm{2}} \\ $$

Commented by abdullah_ff last updated on 22/Nov/21