Question Number 179282 by Shrinava last updated on 27/Oct/22
$$\mathrm{m}^{\mathrm{2}} \:=\:\mathrm{n}\:+\:\mathrm{2} \\ $$$$\mathrm{n}^{\mathrm{2}} \:=\:\mathrm{m}\:+\:\mathrm{2} \\ $$$$\mathrm{4mn}\:−\:\mathrm{m}^{\mathrm{3}} \:−\:\mathrm{n}^{\mathrm{3}} \:=\:?\:\:\:\left(\mathrm{m}\neq\mathrm{n}\right) \\ $$
Answered by Acem last updated on 28/Oct/22
$$\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\:{k}\:=\:\:\mathrm{4}{mn}−\left({m}^{\mathrm{3}} +{n}^{\mathrm{3}} \right) \\ $$$$\:\boldsymbol{{k}}\:=\:\mathrm{4}\boldsymbol{{mn}}−\left(\boldsymbol{{m}}+\boldsymbol{{n}}\right)\:\left(\boldsymbol{{m}}^{\mathrm{2}} −\boldsymbol{{mn}}+\:\boldsymbol{{n}}^{\mathrm{2}} \right)….\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\:{m}^{\mathrm{2}} −{n}^{\mathrm{2}} =\:\left({m}+{n}\right)\left({m}−{n}\right)=\:{n}−{m}\Rightarrow\:\boldsymbol{{m}}+\boldsymbol{{n}}=\:−\mathrm{1}\mid_{{a}} \\ $$$$\:{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\:{m}+{n}+\mathrm{4}\:\Rightarrow\:\boldsymbol{{m}}^{\mathrm{2}} +\boldsymbol{{n}}^{\mathrm{2}} =\:\mathrm{3}\mid_{{b}} \\ $$$$\:\left({m}+{n}\right)^{\mathrm{2}} =\:\mathrm{1}=\:{m}^{\mathrm{2}} +{n}^{\mathrm{2}} +\:\mathrm{2}{mn}=\:\mathrm{3}+\:\mathrm{2}{mn}\Rightarrow\:\boldsymbol{{mn}}=\:−\mathrm{1}\mid_{{c}} \\ $$$$\:{By}\:{compensate}\:{a},\:{b},\:{c}\:\:\:{in}\:\left(\mathrm{1}\right): \\ $$$$\:\:\boldsymbol{{k}}\:=\:\mathrm{0} \\ $$$$ \\ $$
Answered by Frix last updated on 28/Oct/22
![n=m^2 −2 (m^2 −2)^2 =m+2 m^4 −4m^2 −m+2=0 (m−2)(m+1)(m^2 +m−1)=0 m=2 ⇒ n=m [answer would be 0] m=−1 ⇒ n=m [answer would be 6] m=−(1/2)±((√5)/2) ⇒ n=−(1/2)∓((√5)/2) ⇒ answer is 0](https://www.tinkutara.com/question/Q179312.png)
$${n}={m}^{\mathrm{2}} −\mathrm{2} \\ $$$$\left({m}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} ={m}+\mathrm{2} \\ $$$${m}^{\mathrm{4}} −\mathrm{4}{m}^{\mathrm{2}} −{m}+\mathrm{2}=\mathrm{0} \\ $$$$\left({m}−\mathrm{2}\right)\left({m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} +{m}−\mathrm{1}\right)=\mathrm{0} \\ $$$${m}=\mathrm{2}\:\Rightarrow\:{n}={m}\:\left[\mathrm{answer}\:\mathrm{would}\:\mathrm{be}\:\mathrm{0}\right] \\ $$$${m}=−\mathrm{1}\:\Rightarrow\:{n}={m}\:\left[\mathrm{answer}\:\mathrm{would}\:\mathrm{be}\:\mathrm{6}\right] \\ $$$${m}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{n}=−\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$
Commented by Acem last updated on 28/Oct/22
$$\left({m}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} ={m}+\mathrm{2} \\ $$$${m}^{\mathrm{4}} −\mathrm{4}{m}^{\mathrm{2}} \:+\mathrm{4}^{\swarrow} \:−{m}+\mathrm{2}=\mathrm{0}\: \\ $$$$ \\ $$
Commented by Frix last updated on 28/Oct/22
$$\left({m}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} ={m}+\mathrm{2} \\ $$$${m}^{\mathrm{4}} −\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}={m}+\mathrm{2} \\ $$$${m}^{\mathrm{4}} −\mathrm{4}{m}^{\mathrm{2}} −{m}+\mathrm{2}=\mathrm{0} \\ $$
Commented by Acem last updated on 28/Oct/22
$${It}\:{was}\:{optical}\:{illusions}\:\left(:\right. \\ $$
Answered by mr W last updated on 28/Oct/22
![m^3 =mn+2m n^3 =mn+2n m^3 +n^3 =2mn+2(m+n) 4mn−m^3 −n^3 =2[mn−(m+n)] m^2 −n^2 =n−m (m+n+1)(m−n)=0 ⇒m+n=−1 m^2 +n^2 =m+n+4 m^2 +n^2 +2mn−2mn=m+n+4 (m+n)^2 −2mn=m+n+4 (−1)^2 −2mn=−1+4 ⇒mn=−1=m+n ⇒4mn−m^3 −n^3 =0](https://www.tinkutara.com/question/Q179318.png)
$${m}^{\mathrm{3}} ={mn}+\mathrm{2}{m} \\ $$$${n}^{\mathrm{3}} ={mn}+\mathrm{2}{n} \\ $$$${m}^{\mathrm{3}} +{n}^{\mathrm{3}} =\mathrm{2}{mn}+\mathrm{2}\left({m}+{n}\right) \\ $$$$\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{2}\left[{mn}−\left({m}+{n}\right)\right] \\ $$$$ \\ $$$${m}^{\mathrm{2}} −{n}^{\mathrm{2}} ={n}−{m} \\ $$$$\left({m}+{n}+\mathrm{1}\right)\left({m}−{n}\right)=\mathrm{0} \\ $$$$\Rightarrow{m}+{n}=−\mathrm{1} \\ $$$${m}^{\mathrm{2}} +{n}^{\mathrm{2}} ={m}+{n}+\mathrm{4} \\ $$$${m}^{\mathrm{2}} +{n}^{\mathrm{2}} +\mathrm{2}{mn}−\mathrm{2}{mn}={m}+{n}+\mathrm{4} \\ $$$$\left({m}+{n}\right)^{\mathrm{2}} −\mathrm{2}{mn}={m}+{n}+\mathrm{4} \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{mn}=−\mathrm{1}+\mathrm{4} \\ $$$$\Rightarrow{mn}=−\mathrm{1}={m}+{n} \\ $$$$ \\ $$$$\Rightarrow\mathrm{4}{mn}−{m}^{\mathrm{3}} −{n}^{\mathrm{3}} =\mathrm{0} \\ $$
Commented by Shrinava last updated on 29/Oct/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$