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Question Number 152980 by mr W last updated on 03/Sep/21

Commented by mathdanisur last updated on 03/Sep/21

$${\mathrm{a}}^7+{\mathrm{b}}^7+{\mathrm{c}}^7=7\mathrm{abc}{(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}^2$$$${\mathrm{a}}^4+{\mathrm{b}}^4+{\mathrm{c}}^4=2{(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}^2$$$$\mathrm{S}=\frac{7\mathrm{abc}{(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}^2}{\mathrm{abc}\left[2{(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}^2\right]}=\frac{7}{2}=3,5$$

Commented by mr W last updated on 03/Sep/21

$$solutionplease!$$

Commented by mr W last updated on 03/Sep/21

$$find\frac{(a^2+b^2+c^2)(a^5+b^5+c^5)}{a^7+b^7+c^7}=?$$

Commented by mathdanisur last updated on 03/Sep/21

$$=\frac{10}{7}$$

Commented by mathdanisur last updated on 03/Sep/21

$$\mathrm{Thanks}\mathrm{ser}$$$$\frac{{\mathrm{a}}^7+{\mathrm{b}}^7+{\mathrm{c}}^7}{7}=\frac{{\mathrm{a}}^5+{\mathrm{b}}^5+{\mathrm{c}}^5}{5}\frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{2}$$

Commented by mathdanisur last updated on 03/Sep/21

$$\mathrm{a}+\mathrm{b}+\mathrm{c}=0$$$${\mathrm{a}}^{\mathrm{k}}+{\mathrm{b}}^{\mathrm{k}}+{\mathrm{c}}^{\mathrm{k}}={\mathrm{S}}_{\mathrm{k}};\mathrm{k}\in {\mathbb{Z}}^{+}\Rightarrow {\mathrm{S}}_{\mathrm{k}+3}=\mathrm{abc}{\mathrm{S}}_{\mathrm{k}}+\frac{1}{2}{\mathrm{S}}_2{\mathrm{S}}_{\mathrm{k}+1}$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=0;\mathrm{a};\mathrm{b};\mathrm{c}\in \mathrm{I}\mathbb{R}$$$$\Rightarrow {\mathrm{S}}_0=3\Rightarrow {\mathrm{a}}^0+{\mathrm{b}}^0+{\mathrm{c}}^0=3$$$$\Rightarrow {\mathrm{S}}_1=0\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=0$$$$\Rightarrow {\mathrm{S}}_2^2={2\mathrm{S}}_4\Rightarrow {({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2)}^2=2({\mathrm{a}}^4+{\mathrm{b}}^4+{\mathrm{c}}^4)$$$$\Rightarrow {\mathrm{S}}_3=3\mathrm{abc}\Rightarrow {\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3=3\mathrm{abc}$$$$\Rightarrow {6\mathrm{S}}_5={5\mathrm{S}}_2{\mathrm{S}}_3\Rightarrow 6({\mathrm{a}}^5+{\mathrm{b}}^5+{\mathrm{c}}^5)=5({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2)({\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3)$$$$\Rightarrow {6\mathrm{S}}_7={7\mathrm{S}}_3{\mathrm{S}}_4\Rightarrow 6({\mathrm{a}}^7+{\mathrm{b}}^7+{\mathrm{c}}^7)=7({\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3)({\mathrm{a}}^4+{\mathrm{b}}^4+{\mathrm{c}}^4)$$$$\Rightarrow {10\mathrm{S}}_7={7\mathrm{S}}_2{\mathrm{S}}_5\Rightarrow 10({\mathrm{a}}^7+{\mathrm{b}}^7+{\mathrm{c}}^7)=7({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2)({\mathrm{a}}^5+{\mathrm{b}}^5+{\mathrm{c}}^5)$$$$.........$$

Commented by mathdanisur last updated on 03/Sep/21

$$\mathrm{Prove}\mathrm{that}\mathrm{if}\mathrm{a}+\mathrm{b}+\mathrm{c}=0\mathrm{then}:$$$$\frac{{\mathrm{a}}^7+{\mathrm{b}}^7+{\mathrm{c}}^7}{7}=\frac{{\mathrm{a}}^5+{\mathrm{b}}^5+{\mathrm{c}}^5}{5}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{2}$$

Commented by mr W last updated on 03/Sep/21

$$thanksalotsir!$$

Commented by mr W last updated on 03/Sep/21

$$but$$$$\frac{{\mathrm{a}}^7+{\mathrm{b}}^7+{\mathrm{c}}^7}{7}=\frac{{\mathrm{a}}^5+{\mathrm{b}}^5+{\mathrm{c}}^5}{5}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{2}$$$$isnottrue.$$

Commented by Tawa11 last updated on 04/Sep/21

$$\mathrm{Nice}$$

Answered by ajfour last updated on 03/Sep/21

$$a=2p,b=c=-p$$$$\frac{(128-2)}{2(16+2)}=\frac{126}{36}=\frac{7}{2}$$

Commented by mr W last updated on 03/Sep/21

$$thankssir!$$

Answered by mr W last updated on 04/Sep/21

$$p_k=a^k+b^k+c^k$$$$e_1=a+b+c$$$$e_2=ab+bc+ca$$$$e_3=abc$$$$$$$$p_1=e_1=0$$$$p_2=e_1{p}_1-2e_2=-2e_2$$$$p_3=e_1{p}_2-e_2{p}_1+3e_3=3e_3$$$$p_4=e_1{p}_3-e_2{p}_2+e_3{p}_1=2e_2^2$$$$p_5=e_1{p}_4-e_2{p}_3+e_3{p}_2=-5e_2{e}_3$$$$p_6=e_1{p}_5-e_2{p}_4+e_3{p}_3=-2e_2^3+3e_3^2$$$$p_7=e_1{p}_6-e_2{p}_5+e_3{p}_4=7e_2^2{e}_3$$$$\Rightarrow \frac{a^7+b^7+c^7}{abc(a^4+b^4+c^4)}=\frac{p^7}{e_3{p}_4}=\frac{7e_2^2{e}_3}{e_3\times 2e_2^2}=\frac{7}{2}$$$$\frac{a^7+b^7+c^7}{7}=e_2^2{e}_3=\frac{-2e_2}{2}\times \frac{-5e_2{e}_3}{5}=\frac{p_2}{2}\times \frac{p_5}{5}$$$$\Rightarrow \frac{a^7+b^7+c^7}{7}=\frac{a^2+b^2+c^2}{2}\times \frac{a^5+b^5+c^5}{5}$$

Commented by mathdanisur last updated on 04/Sep/21

$$\mathrm{thanks}\mathrm{ser}$$

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