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if-m-3-2p-3-3mn-a-3-b-3-p-3-and-a-2-b-2-n-then-prove-that-a-b-m-




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

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