Question Number 70103 by Shamim last updated on 01/Oct/19
$$\mathrm{if}\:\mathrm{m}^{\mathrm{3}} +\mathrm{2p}^{\mathrm{3}} =\mathrm{3mn},\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\mathrm{p}^{\mathrm{3}} \:\mathrm{and} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{n}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{a}+\mathrm{b}=\mathrm{m}. \\ $$
Answered by mind is power last updated on 01/Oct/19
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{ab}\right) \\ $$$$\mathrm{x}=\mathrm{a}+\mathrm{b} \\ $$$$\Rightarrow{x}\left({n}−{ab}\right)={p}^{\mathrm{3}} \\ $$$${ab}=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({a}+{b}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right\}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} −{n}\right) \\ $$$$\Rightarrow{p}^{\mathrm{3}} ={x}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} −{n}\right)\right) \\ $$$$\Rightarrow\mathrm{2}{p}^{\mathrm{3}} =\mathrm{2}{nx}−{x}^{\mathrm{3}} +{nx} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{2}{p}^{\mathrm{3}} =\mathrm{3}{nx}...{E} \\ $$$${we}\:{have}\:{m}\:{solution}\:{of}\:..{E} \\ $$$${cause}\:{m}^{\mathrm{3}} +\mathrm{2}{p}^{\mathrm{3}} =\mathrm{3}{mn} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{nx}+\mathrm{2}{p}^{\mathrm{3}} =\mathrm{0} \\ $$$${m}^{\mathrm{3}} −\mathrm{3}{mn}+\mathrm{2}{p}^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow\left({x}−{m}\right)\left({x}^{\mathrm{2}} +{m}^{\mathrm{2}} −{mx}−\mathrm{3}{n}\right)=\mathrm{0} \\ $$$${x}={m} \\ $$$${or}\:{x}^{\mathrm{2}} −{mx}+{m}^{\mathrm{2}} −\mathrm{3}{n}=\mathrm{0} \\ $$$${we}\:{have}\:{other}\:{case}\:{but}\:{depend}\:{if}\:{we}\:\:{solve}\:{in}/\mathbb{R}\:{or}\:\mathbb{C} \\ $$$$ \\ $$
Answered by MJS last updated on 01/Oct/19
![m^3 +2p^3 =3mn ⇒ p^3 =(m/2)(3n−m^2 ) (1) a+b=x (2) a^2 +b^2 =n (3) a^3 +b^3 =(m/2)(3n−m^2 ) put a=α−(√β)∧b=α+(√β) (1) 2α=x (2) 2α^2 +2β=n (3) 2α^3 +6αβ=(m/2)(3n−m^2 ) (1) ⇒ α=(x/2) (2) ⇒ β=(n/2)−(x^2 /4) (3) ⇒ β=((mn)/(2x))−(m^3 /(6x))−(x^2 /(12)) ⇒ (n/2)−(x^2 /4)=((mn)/(2x))−(m^3 /(6x))−(x^2 /(12)) ⇔ x^3 −3nx−m(m^2 −3n)=0 try x=±m∨x=±(m^2 −3n)∨x=±m(m^2 −3n) ⇒ x_1 =m [(x−m)(x^2 +mx+m^2 −3n)=0 ⇒ x_(2, 3) =−(m/2)±((√(12n−3m^2 ))/2)] ⇒ a+b=m](Q70107.png)
$${m}^{\mathrm{3}} +\mathrm{2}{p}^{\mathrm{3}} =\mathrm{3}{mn}\:\Rightarrow\:{p}^{\mathrm{3}} =\frac{{m}}{\mathrm{2}}\left(\mathrm{3}{n}−{m}^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{1}\right)\:\:{a}+{b}={x} \\ $$$$\left(\mathrm{2}\right)\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={n} \\ $$$$\left(\mathrm{3}\right)\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\frac{{m}}{\mathrm{2}}\left(\mathrm{3}{n}−{m}^{\mathrm{2}} \right) \\ $$$$\mathrm{put}\:{a}=\alpha−\sqrt{\beta}\wedge{b}=\alpha+\sqrt{\beta} \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{2}\alpha={x} \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{2}\beta={n} \\ $$$$\left(\mathrm{3}\right)\:\:\mathrm{2}\alpha^{\mathrm{3}} +\mathrm{6}\alpha\beta=\frac{{m}}{\mathrm{2}}\left(\mathrm{3}{n}−{m}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\:\Rightarrow\:\alpha=\frac{{x}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\:\Rightarrow\:\beta=\frac{{n}}{\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\left(\mathrm{3}\right)\:\:\Rightarrow\:\beta=\frac{{mn}}{\mathrm{2}{x}}−\frac{{m}^{\mathrm{3}} }{\mathrm{6}{x}}−\frac{{x}^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\Rightarrow \\ $$$$\frac{{n}}{\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}=\frac{{mn}}{\mathrm{2}{x}}−\frac{{m}^{\mathrm{3}} }{\mathrm{6}{x}}−\frac{{x}^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\Leftrightarrow \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{nx}−{m}\left({m}^{\mathrm{2}} −\mathrm{3}{n}\right)=\mathrm{0} \\ $$$$\mathrm{try}\:{x}=\pm{m}\vee{x}=\pm\left({m}^{\mathrm{2}} −\mathrm{3}{n}\right)\vee{x}=\pm{m}\left({m}^{\mathrm{2}} −\mathrm{3}{n}\right) \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} ={m} \\ $$$$\:\:\:\:\:\left[\left({x}−{m}\right)\left({x}^{\mathrm{2}} +{mx}+{m}^{\mathrm{2}} −\mathrm{3}{n}\right)=\mathrm{0}\right. \\ $$$$\left.\:\:\:\:\:\:\:\Rightarrow\:{x}_{\mathrm{2},\:\mathrm{3}} =−\frac{{m}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{12}{n}−\mathrm{3}{m}^{\mathrm{2}} }}{\mathrm{2}}\right] \\ $$$$ \\ $$$$\Rightarrow\:{a}+{b}={m} \\ $$