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Question Number 155801 by cortano last updated on 05/Oct/21
{ ((m^2 =n+2)),((n^2 =m+2)) :} ⇒m≠n 4mn−m^3 −n^3 =?
$$\:\begin{cases}{\mathrm{m}^{\mathrm{2}} =\mathrm{n}+\mathrm{2}}\\{\mathrm{n}^{\mathrm{2}} =\mathrm{m}+\mathrm{2}}\end{cases}\:\Rightarrow\mathrm{m}\neq\mathrm{n} \\ $$$$\:\mathrm{4mn}−\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} =?\: \\ $$
Answered by Rasheed.Sindhi last updated on 05/Oct/21
{ ((m^2 =n+2...(i))),((n^2 =m+2...(ii))) :} ∧ m≠n 4mn−m^3 −n^3 =? (i)−(ii): m^2 −n^2 =n−m (m−n)(m+n)=−(m−n) m+n=−1 (m+n)^3 =(−1)^3 m^3 +n^3 +3mn(m+n)=−1 −m^3 −n^3 −3mn(−1)=1 3mn−m^3 −n^3 =1............A (i)×(ii): m^2 n^2 =mn+2(m+n)+4 (mn)^2 −mn−2(−1)−4=0 (mn)^2 −mn−2=0 (mn+1)(mn−2)=0 mn=−1,mn=2 A⇒3mn−m^3 −n^3 +mn=1+mn { ((4mn−m^3 −n^3 =1+(−1)=0)),((4mn−m^3 −n^3 =1+(2)=3)) :}
$$\:\begin{cases}{\mathrm{m}^{\mathrm{2}} =\mathrm{n}+\mathrm{2}…\left({i}\right)}\\{\mathrm{n}^{\mathrm{2}} =\mathrm{m}+\mathrm{2}…\left({ii}\right)}\end{cases}\:\:\wedge\:\:\mathrm{m}\neq\mathrm{n}\:\: \\ $$$$\:\mathrm{4mn}−\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} =?\: \\ $$$$\left({i}\right)−\left({ii}\right):\:{m}^{\mathrm{2}} −{n}^{\mathrm{2}} ={n}−{m} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({m}−{n}\right)\left({m}+{n}\right)=−\left({m}−{n}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{m}+{n}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left({m}+{n}\right)^{\mathrm{3}} =\left(−\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:{m}^{\mathrm{3}} +{n}^{\mathrm{3}} +\mathrm{3}{mn}\left({m}+{n}\right)=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} −\mathrm{3}{mn}\left(−\mathrm{1}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{3}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{1}…………{A} \\ $$$$\left({i}\right)×\left({ii}\right):\:{m}^{\mathrm{2}} {n}^{\mathrm{2}} ={mn}+\mathrm{2}\left({m}+{n}\right)+\mathrm{4}\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\left({mn}\right)^{\mathrm{2}} −{mn}−\mathrm{2}\left(−\mathrm{1}\right)−\mathrm{4}=\mathrm{0}\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\left({mn}\right)^{\mathrm{2}} −{mn}−\mathrm{2}=\mathrm{0}\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\left({mn}+\mathrm{1}\right)\left({mn}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:{mn}=−\mathrm{1},{mn}=\mathrm{2} \\ $$$${A}\Rightarrow\mathrm{3}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} +{mn}=\mathrm{1}+{mn} \\ $$$$\begin{cases}{\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{1}+\left(−\mathrm{1}\right)=\mathrm{0}}\\{\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{1}+\left(\mathrm{2}\right)=\mathrm{3}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\: \\ $$
Commented by cortano last updated on 05/Oct/21
sorry sir. not correct. if 4mn−m^3 −n^3 = 0 it follows that m=n=2 .
$$\mathrm{sorry}\:\mathrm{sir}.\:\mathrm{not}\:\mathrm{correct}. \\ $$$$\mathrm{if}\:\mathrm{4mn}−\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} =\:\mathrm{0}\:\mathrm{it} \\ $$$$\:\mathrm{follows}\:\mathrm{that}\:\mathrm{m}=\mathrm{n}=\mathrm{2}\:. \\ $$
Commented by Rasheed.Sindhi last updated on 05/Oct/21
• 4mn−m^3 −n^3 =0⇏m=n=2 •What about other answer: 4mn−m^3 −n^3 =1+(2)=3 If it′s also incorrect, what′s right answer?
$$\:\bullet\:\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{0}\nRightarrow{m}={n}=\mathrm{2} \\ $$$$\:\bullet{What}\:{about}\:{other}\:{answer}: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{1}+\left(\mathrm{2}\right)=\mathrm{3} \\ $$$${If}\:{it}'{s}\:{also}\:{incorrect},\:{what}'{s}\:{right} \\ $$$${answer}? \\ $$
Commented by mr W last updated on 07/Oct/21
cortano sir: both rasheed sir and i have given a detailed working for the result zero. you said this result is wrong, but you can not tell us where it is wrong in our working. i think nobody can tell because both working and result are correct. your answer that the answer is not zero is clearly wrong. in fact to get the result we don′t need to calculate the values of m and n. it′s ok you did. you got the correct values for m, n: n,m=((−1±(√5))/2) ✓. but with theses values you should and must get 4mm−m^3 −n^3 =0. if not, you should buy a new calculator.
$${cortano}\:{sir}: \\ $$$${both}\:{rasheed}\:{sir}\:{and}\:{i}\:{have}\:{given}\:{a} \\ $$$${detailed}\:{working}\:{for}\:{the}\:{result}\:{zero}. \\ $$$${you}\:{said}\:{this}\:{result}\:{is}\:{wrong},\:{but} \\ $$$${you}\:{can}\:{not}\:{tell}\:{us}\:{where}\:{it}\:{is}\:{wrong} \\ $$$${in}\:{our}\:{working}.\:{i}\:{think}\:{nobody}\:{can}\:{tell}\: \\ $$$${because}\:{both}\:{working}\:{and}\:{result} \\ $$$${are}\:{correct}. \\ $$$${your}\:{answer}\:{that}\:{the}\:{answer}\:{is}\:{not} \\ $$$${zero}\:{is}\:{clearly}\:{wrong}.\:{in}\:{fact}\:{to}\:{get} \\ $$$${the}\:{result}\:{we}\:{don}'{t}\:{need}\:{to}\:{calculate} \\ $$$${the}\:{values}\:{of}\:{m}\:{and}\:{n}.\:{it}'{s}\:{ok}\:{you}\:{did}. \\ $$$${you}\:{got}\:{the}\:{correct}\:{values}\:{for}\:{m},\:{n}: \\ $$$${n},{m}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\checkmark. \\ $$$${but}\:{with}\:{theses}\:{values}\:{you}\:{should}\:{and} \\ $$$${must}\:{get}\:\mathrm{4}{mm}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{0}.\:{if}\:{not}, \\ $$$${you}\:{should}\:{buy}\:{a}\:{new}\:{calculator}. \\ $$
Commented by Rasheed.Sindhi last updated on 07/Oct/21
مھرباني منھنجا سائين!
Commented by Rasheed.Sindhi last updated on 07/Oct/21
Sir mr W ,how can we reject mn=2 in my above anwser?
$${Sir}\:{mr}\:{W}\:,{how}\:{can}\:{we}\:{reject}\:\mathrm{mn}=\mathrm{2} \\ $$$${in}\:{my}\:{above}\:{anwser}? \\ $$
Commented by mr W last updated on 07/Oct/21
answer 0 is correct!
$${answer}\:\mathrm{0}\:{is}\:{correct}! \\ $$
Commented by cortano last updated on 07/Oct/21
oo yes m≠n≠2 but m=n=((±(√5)−1)/2) then 4mn−m^3 −n^3 =0. sorry sir
$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{m}\neq\mathrm{n}\neq\mathrm{2}\:\mathrm{but}\:\mathrm{m}=\mathrm{n}=\frac{\pm\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{then}\:\mathrm{4mn}−\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} =\mathrm{0}. \\ $$$$\mathrm{sorry}\:\mathrm{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 07/Oct/21
But you have given condition mr cortano that m≠n,then how m=n=((±(√5)−1)/2) ?
$$\mathcal{B}{ut}\:{you}\:{have}\:{given}\:{condition}\:{mr}\:{cortano} \\ $$$${that}\:{m}\neq{n},{then}\:{how}\:\mathrm{m}=\mathrm{n}=\frac{\pm\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:? \\ $$
Commented by Rasheed.Sindhi last updated on 07/Oct/21
m=n=((±(√5)−1)/2) { ((m^2 =n+2⇒((((√5)−1)/2))^2 =^(?) (((√5)−1)/2)+2)),((n^2 =m+2⇒((((√5)−1)/2))^2 =^(?) (((√5)−1)/2)+2)) :} { ((m^2 =n+2⇒((((√5)−1)/2))^2 =^(?) (((√5)−1)/2)+2)),((n^2 =m+2⇒((((√5)−1)/2))^2 =^(?) (((√5)−1)/2)+2)) :} ((5+1−2(√5))/4)=^(?) (((√5)−1+4)/2) ((3−(√5))/2)≠(((√5)+3)/2) Don′t satisfied by m=n=((±(√5)−1)/2)
$$\mathrm{m}=\mathrm{n}=\frac{\pm\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\begin{cases}{\mathrm{m}^{\mathrm{2}} =\mathrm{n}+\mathrm{2}\Rightarrow\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \overset{?} {=}\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}+\mathrm{2}}\\{\mathrm{n}^{\mathrm{2}} =\mathrm{m}+\mathrm{2}\Rightarrow\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \overset{?} {=}\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}+\mathrm{2}}\end{cases}\: \\ $$$$\begin{cases}{\mathrm{m}^{\mathrm{2}} =\mathrm{n}+\mathrm{2}\Rightarrow\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \overset{?} {=}\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}+\mathrm{2}}\\{\mathrm{n}^{\mathrm{2}} =\mathrm{m}+\mathrm{2}\Rightarrow\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \overset{?} {=}\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}+\mathrm{2}}\end{cases}\: \\ $$$$\frac{\mathrm{5}+\mathrm{1}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}\overset{?} {=}\frac{\sqrt{\mathrm{5}}−\mathrm{1}+\mathrm{4}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\neq\frac{\sqrt{\mathrm{5}}+\mathrm{3}}{\mathrm{2}} \\ $$$${Don}'{t}\:{satisfied}\:{by}\:{m}={n}=\frac{\pm\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 07/Oct/21
rasheed sir: with m+n=−1 m=−(n+1) P=mn=−(1+n)n=(1/4)−(n+(1/2))^2 ≤(1/4) therefore mn=2>(1/4) must be rejected.
$${rasheed}\:{sir}: \\ $$$${with}\:{m}+{n}=−\mathrm{1} \\ $$$${m}=−\left({n}+\mathrm{1}\right) \\ $$$${P}={mn}=−\left(\mathrm{1}+{n}\right){n}=\frac{\mathrm{1}}{\mathrm{4}}−\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${therefore}\:{mn}=\mathrm{2}>\frac{\mathrm{1}}{\mathrm{4}}\:{must}\:{be}\:{rejected}. \\ $$
Commented by cortano last updated on 07/Oct/21
Commented by cortano last updated on 07/Oct/21
clear the answer is not zero
$$\mathrm{clear}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{not}\:\mathrm{zero} \\ $$
Commented by Rasheed.Sindhi last updated on 07/Oct/21
THANKS-a-lot Sir mr W!
$$\mathcal{THANKS}-{a}-{lot}\:\:\mathcal{S}{ir}\:{mr}\:\mathcal{W}! \\ $$
Commented by cortano last updated on 07/Oct/21
the correct answer is 9+(√5) not zero. if 4mn−m^3 −n^3 =0 it follows that m=n=2. my solution { ((m=(((√5)−1)/2))),((n=((−(√5)−1)/2))) :}
$$\mathrm{the}\:\mathrm{correct}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{9}+\sqrt{\mathrm{5}} \\ $$$$\mathrm{not}\:\mathrm{zero}.\:\mathrm{if}\:\mathrm{4mn}−\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{m}=\mathrm{n}=\mathrm{2}. \\ $$$$\mathrm{my}\:\mathrm{solution}\:\begin{cases}{\mathrm{m}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}}\\{\mathrm{n}=\frac{−\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$
Commented by mr W last updated on 07/Oct/21
why don′t you calculate 4mn−m^3 −n^3 simply with your values: { ((m=(((√5)−1)/2))),((n=((−(√5)−1)/2))) :} here i check for you: 4mn=4×(((√5)−1)/2)×((−(√5)−1)/2) =−((√5)−1)((√5)+1) =−[((√5))^2 −1^2 ] =−(5−1) =−(4) =−4 ≠5+(√5) besides 4mn−m^3 −n^3 =0 doesn′t lead to m=n=2 only! in fact 4mn−m^3 −n^3 =0 has infinite solutions. m=n=2 is only one of infinite many! [edited]
$${why}\:{don}'{t}\:{you}\:{calculate}\:\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} \\ $$$${simply}\:{with}\:{your}\:{values}: \\ $$$$\begin{cases}{\mathrm{m}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}}\\{\mathrm{n}=\frac{−\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$$${here}\:{i}\:{check}\:{for}\:{you}: \\ $$$$\mathrm{4}{mn}=\mathrm{4}×\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}×\frac{−\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$=−\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\left(\sqrt{\mathrm{5}}+\mathrm{1}\right) \\ $$$$=−\left[\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right] \\ $$$$=−\left(\mathrm{5}−\mathrm{1}\right) \\ $$$$=−\left(\mathrm{4}\right) \\ $$$$=−\mathrm{4} \\ $$$$\neq\mathrm{5}+\sqrt{\mathrm{5}} \\ $$$$ \\ $$$${besides}\:\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{0}\:{doesn}'{t} \\ $$$${lead}\:{to}\:{m}={n}=\mathrm{2}\:{only}!\:{in}\:{fact} \\ $$$$\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{0}\:{has}\:{infinite} \\ $$$${solutions}.\:{m}={n}=\mathrm{2}\:{is}\:{only}\:{one}\:{of} \\ $$$${infinite}\:{many}! \\ $$$$ \\ $$$$\left[{edited}\right] \\ $$
Answered by john_santu last updated on 07/Oct/21
(1)m^2 +n^2 =m+n+4 (2)m^3 +n^3 =(m+n)(m^2 +n^2 −mn) =(m+n)(m+n−mn+4) (3)(m^2 −2)^2 =m+2 ⇒m^4 −4m^2 −m+2=0 ⇒(m+1)(m−2)(m^2 +m−1)=0 m=−1 ∧m=2 rejected since m≠n ⇒m=((−1±(√5))/2) ∧n=((∓2(√5)−2)/4)
$$\left(\mathrm{1}\right){m}^{\mathrm{2}} +{n}^{\mathrm{2}} ={m}+{n}+\mathrm{4} \\ $$$$\left(\mathrm{2}\right){m}^{\mathrm{3}} +{n}^{\mathrm{3}} =\left({m}+{n}\right)\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} −{mn}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left({m}+{n}\right)\left({m}+{n}−{mn}+\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right)\left({m}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} ={m}+\mathrm{2} \\ $$$$\Rightarrow{m}^{\mathrm{4}} −\mathrm{4}{m}^{\mathrm{2}} −{m}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({m}+\mathrm{1}\right)\left({m}−\mathrm{2}\right)\left({m}^{\mathrm{2}} +{m}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:{m}=−\mathrm{1}\:\wedge{m}=\mathrm{2}\:{rejected}\:{since}\:{m}\neq{n} \\ $$$$\Rightarrow{m}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\wedge{n}=\frac{\mp\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by cortano last updated on 05/Oct/21
thanks
$$\mathrm{thanks} \\ $$
Answered by mr W last updated on 07/Oct/21
m^2 −n^2 =n−m (m+n)(m−n)=n−m ⇒m+n=−1 m^2 +n^2 =m+n+4 (m+n)^2 −2mn=m+n+4 1−2mn=−1+4 ⇒mn=−1 { m,n are roots of z^2 +z−1=0 ⇒m,n=z=((−1±(√5))/2) } (m+n)^3 =m^3 +n^3 +3mn(m+n) (−1)^3 =m^3 +n^3 +3(−1)(−1) ⇒m^3 +n^3 =−4 4mn−m^3 −n^3 =4(−1)−(−4)=0
$${m}^{\mathrm{2}} −{n}^{\mathrm{2}} ={n}−{m} \\ $$$$\left({m}+{n}\right)\left({m}−{n}\right)={n}−{m} \\ $$$$\Rightarrow{m}+{n}=−\mathrm{1} \\ $$$${m}^{\mathrm{2}} +{n}^{\mathrm{2}} ={m}+{n}+\mathrm{4} \\ $$$$\left({m}+{n}\right)^{\mathrm{2}} −\mathrm{2}{mn}={m}+{n}+\mathrm{4} \\ $$$$\mathrm{1}−\mathrm{2}{mn}=−\mathrm{1}+\mathrm{4} \\ $$$$\Rightarrow{mn}=−\mathrm{1} \\ $$$$\left\{\right. \\ $$$${m},{n}\:{are}\:{roots}\:{of}\:{z}^{\mathrm{2}} +{z}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{m},{n}={z}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\: \\ $$$$\left.\right\} \\ $$$$\left({m}+{n}\right)^{\mathrm{3}} ={m}^{\mathrm{3}} +{n}^{\mathrm{3}} +\mathrm{3}{mn}\left({m}+{n}\right) \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{3}} ={m}^{\mathrm{3}} +{n}^{\mathrm{3}} +\mathrm{3}\left(−\mathrm{1}\right)\left(−\mathrm{1}\right) \\ $$$$\Rightarrow{m}^{\mathrm{3}} +{n}^{\mathrm{3}} =−\mathrm{4} \\ $$$$ \\ $$$$\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{4}\left(−\mathrm{1}\right)−\left(−\mathrm{4}\right)=\mathrm{0} \\ $$

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