Question Number 71499 by petrochengula last updated on 16/Oct/19
Commented by petrochengula last updated on 16/Oct/19
$$\mathrm{help}\:\mathrm{2a} \\ $$
Answered by MJS last updated on 16/Oct/19
![cosh x =(1/2)(e^x +e^(−x) ) cosh 2x =(1/2)(e^(2x) +e^(−2x) ) ((cosh 2x)/(cosh x))=e^x +e^(−x) −(2/(e^x +e^(−x) ))=2cosh x −(1/(cosh x)) ⇒ ⇒ cosh 2x =2cosh^2 x −1 cosh 3x =(1/2)(e^(3x) +e^(−3x) ) ((cosh 3x)/(cosh x))=e^(2x) +e^(−2x) −1=2cosh 2x −1= =4cosh^2 x −3 ⇒ cosh 3x =4cosh^3 x −3cosh x acosh 3x +bcosh 2x =0 [cosh x =c] 4ac^3 +2bc^2 −3ac−b=0 c^3 +(b/(2a))c^2 −(3/4)c−(b/(4a))=0 (c−γ_1 )(c−γ_2 )(c−γ_3 )= =c^3 −(γ_1 +γ_2 +γ_3 )c^2 +(γ_1 γ_2 +γ_1 γ_3 +γ_2 γ_3 )c−γ_1 γ_2 γ_3 ⇒ γ_1 γ_2 +γ_1 γ_3 +γ_2 γ_3 =−(3/4) independent of a, b](Q71510.png)
$$\mathrm{cosh}\:{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} \right) \\ $$$$\mathrm{cosh}\:\mathrm{2}{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{2}{x}} +\mathrm{e}^{−\mathrm{2}{x}} \right) \\ $$$$\frac{\mathrm{cosh}\:\mathrm{2}{x}}{\mathrm{cosh}\:{x}}=\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} −\frac{\mathrm{2}}{\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} }=\mathrm{2cosh}\:{x}\:−\frac{\mathrm{1}}{\mathrm{cosh}\:{x}}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{cosh}\:\mathrm{2}{x}\:=\mathrm{2cosh}^{\mathrm{2}} \:{x}\:−\mathrm{1} \\ $$$$\mathrm{cosh}\:\mathrm{3}{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{3}{x}} +\mathrm{e}^{−\mathrm{3}{x}} \right) \\ $$$$\frac{\mathrm{cosh}\:\mathrm{3}{x}}{\mathrm{cosh}\:{x}}=\mathrm{e}^{\mathrm{2}{x}} +\mathrm{e}^{−\mathrm{2}{x}} −\mathrm{1}=\mathrm{2cosh}\:\mathrm{2}{x}\:−\mathrm{1}= \\ $$$$=\mathrm{4cosh}^{\mathrm{2}} \:{x}\:−\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{cosh}\:\mathrm{3}{x}\:=\mathrm{4cosh}^{\mathrm{3}} \:{x}\:−\mathrm{3cosh}\:{x} \\ $$$$ \\ $$$${a}\mathrm{cosh}\:\mathrm{3}{x}\:+{b}\mathrm{cosh}\:\mathrm{2}{x}\:=\mathrm{0} \\ $$$$\:\:\:\:\:\left[\mathrm{cosh}\:{x}\:={c}\right] \\ $$$$\mathrm{4}{ac}^{\mathrm{3}} +\mathrm{2}{bc}^{\mathrm{2}} −\mathrm{3}{ac}−{b}=\mathrm{0} \\ $$$${c}^{\mathrm{3}} +\frac{{b}}{\mathrm{2}{a}}{c}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{4}}{c}−\frac{{b}}{\mathrm{4}{a}}=\mathrm{0} \\ $$$$\left({c}−\gamma_{\mathrm{1}} \right)\left({c}−\gamma_{\mathrm{2}} \right)\left({c}−\gamma_{\mathrm{3}} \right)= \\ $$$$={c}^{\mathrm{3}} −\left(\gamma_{\mathrm{1}} +\gamma_{\mathrm{2}} +\gamma_{\mathrm{3}} \right){c}^{\mathrm{2}} +\left(\gamma_{\mathrm{1}} \gamma_{\mathrm{2}} +\gamma_{\mathrm{1}} \gamma_{\mathrm{3}} +\gamma_{\mathrm{2}} \gamma_{\mathrm{3}} \right){c}−\gamma_{\mathrm{1}} \gamma_{\mathrm{2}} \gamma_{\mathrm{3}} \\ $$$$\Rightarrow\:\gamma_{\mathrm{1}} \gamma_{\mathrm{2}} +\gamma_{\mathrm{1}} \gamma_{\mathrm{3}} +\gamma_{\mathrm{2}} \gamma_{\mathrm{3}} =−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{independent}\:\mathrm{of}\:{a},\:{b} \\ $$
Commented by petrochengula last updated on 16/Oct/19
$$\mathrm{thanks} \\ $$$$ \\ $$