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Question Number 74970 by mr W last updated on 05/Dec/19

$$x+y+z=1$$$$x^2+y^2+z^2=2$$$$x^3+y^3+z^3=3$$$$$$$$find$$$$x^4+y^4+z^4=?$$$$x^5+y^5+z^5=?$$$$x^6+y^6+z^6=?$$$$......$$$$x^n+y^n+z^n=?$$

Commented by TawaTawa last updated on 05/Dec/19

$$\mathrm{Wow},\mathrm{i}\mathrm{will}\mathrm{love}\mathrm{to}\mathrm{see}\mathrm{the}\mathrm{answer}.$$

Commented by mr W last updated on 05/Dec/19

Commented by mr W last updated on 05/Dec/19

Commented by mr W last updated on 05/Dec/19

https://en.m.wikipedia.org/wiki/Newton%27s_identities

Commented by mr W last updated on 05/Dec/19

$$p_k=x^k+y^k+z^k$$$$e_1=x+y+z$$$$e_2=xy+yz+zx$$$$e_3=xyz$$$$e_{i⩾4}=0$$$$e_1=p_1=1$$$$2e_2=e_1{p}_1-p_2=1-2=-1\Rightarrow e_2=-\frac{1}{2}$$$$3e_3=e_2{p}_1-e_1{p}_2+p_3=-\frac{1}{2}-2+3=\frac{1}{2}\Rightarrow e_3=\frac{1}{6}$$$$0=e_3{p}_1-e_2{p}_2+e_1{p}_3-p_4\Rightarrow p_4=\frac{1}{6}+1+3=\frac{25}{6}$$$$0=-e_3{p}_2+e_2{p}_3-e_1{p}_4+p_5\Rightarrow p_5=\frac{2}{6}+\frac{3}{2}+\frac{25}{6}=6$$$$0=e_3{p}_3-e_2{p}_4+e_1{p}_5-p_6\Rightarrow p_6=\frac{3}{6}+\frac{25}{12}+6=\frac{103}{12}$$$$0=-e_3{p}_4+e_2{p}_5-e_1{p}_6+p_7\Rightarrow p_7=\frac{25}{36}+\frac{6}{2}+\frac{103}{12}=\frac{221}{18}$$$$......$$$$p_n=p_{n-1}+\frac{p_{n-2}}{2}+\frac{p_{n-3}}{6}$$$$......$$$$i.e.$$$$x^4+y^4+z^4=\frac{25}{6}$$$$x^5+y^5+z^5=6$$$$x^6+y^6+z^6=\frac{103}{12}$$$$x^7+y^7+z^7=\frac{221}{18}$$$$x^8+y^8+z^8=\frac{1265}{72}$$$$......$$

Commented by TawaTawa last updated on 05/Dec/19

$$\mathrm{Wow},\mathrm{i}\mathrm{will}\mathrm{study}\mathrm{them}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sirs}$$

Commented by TawaTawa last updated on 05/Dec/19

$$\mathrm{The}\mathrm{only}\mathrm{thing}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{is}\mathrm{how}\mathrm{to}\mathrm{get}$$$${2\mathrm{e}}_2,{3\mathrm{e}}_3,{4\mathrm{e}}_4,{5\mathrm{e}}_5,\mathrm{etc}...$$$$\mathrm{Please}\mathrm{help}$$

Commented by mind is power last updated on 05/Dec/19

$$\mathrm{for}3\mathrm{variable}$$$${\mathrm{S}}_2=\mathrm{xy}+\mathrm{yz}+\mathrm{zx}$$$${\mathrm{S}}_3=\mathrm{xyz}$$$${\mathrm{S}}_4=0$$$$\mathrm{Why}{\mathrm{S}}_4=0$$$${\mathrm{P}}_4={\mathrm{x}}^4+{\mathrm{y}}^4+{\mathrm{z}}^4={\mathrm{x}}^4+{\mathrm{y}}^4+{\mathrm{z}}^4+0^4$$$$\mathrm{if}\mathrm{we}\mathrm{applie}\mathrm{too}(\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d})=(\mathrm{x},\mathrm{y},\mathrm{z},0)$$$$\mathrm{samme}\mathrm{definitiln}$$$${\mathrm{S}}_1(\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d})=\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=\mathrm{x}+\mathrm{y}+\mathrm{z}+0$$$${\mathrm{S}}_2(\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d})=\mathrm{ab}+\mathrm{ac}+\mathrm{ad}+\mathrm{bc}+\mathrm{bd}+\mathrm{cd}$$$$\mathrm{since}\mathrm{d}=0,\mathrm{a}=\mathrm{x},\mathrm{b}=\mathrm{y},\mathrm{c}=\mathrm{z}\Rightarrow$$$${\mathrm{S}}_2=\mathrm{xy}+\mathrm{xz}+\mathrm{zy}$$$${\mathrm{s}}_3=\mathrm{xyz}$$$${\mathrm{s}}_4=\mathrm{abcd}=\mathrm{xyz}.0=0$$$$$$

Commented by TawaTawa last updated on 05/Dec/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{i}\mathrm{understand}\mathrm{this}$$$$\mathrm{but}\mathrm{i}\mathrm{mean}:\mathrm{how}\mathrm{to}\mathrm{get}$$$${2\mathrm{e}}_2={\mathrm{e}}_1{\mathrm{p}}_1-{\mathrm{p}}_2$$$${3\mathrm{e}}_3={\mathrm{e}}_2{\mathrm{p}}_1-{\mathrm{e}}_1{\mathrm{p}}_2+{\mathrm{p}}_3$$$${4\mathrm{e}}_4=....$$$$\mathrm{etc}..$$

Commented by mind is power last updated on 05/Dec/19

$${(\mathrm{x}+\mathrm{y}+\mathrm{z})}^2={\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2+2\mathrm{xy}+2\mathrm{xz}+2\mathrm{zy}$$$$\Rightarrow {\mathrm{p}}_1^2={\mathrm{p}}_2+{2\mathrm{e}}_2\Rightarrow {2\mathrm{e}}_2={\mathrm{p}}_1^2-{\mathrm{p}}_2$$$$(\mathrm{xy}+\mathrm{xz}+\mathrm{zx})(\mathrm{x}+\mathrm{y}+\mathrm{z})-(\mathrm{x}+\mathrm{y}+\mathrm{z})({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2)+{\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3$$$$=3\mathrm{xyz}={3\mathrm{e}}_3\mathrm{just}\mathrm{devllope}\mathrm{and}\mathrm{simplifaction}$$

Commented by TawaTawa last updated on 05/Dec/19

$$\mathrm{I}\mathrm{understand}\mathrm{now}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by mind is power last updated on 05/Dec/19

$$\mathrm{withe}\mathrm{pleasur}\mathrm{Sir}$$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{Great}.\mathrm{I}\mathrm{found}\mathrm{it}.$$

Answered by MJS last updated on 05/Dec/19

$$x=p-\sqrt{q}$$$$y=p+\sqrt{q}$$$$z=1-x-y=1-2p$$$$\left(1\right)\mathrm{true}$$$$\left(2\right)6p^2+2q-4p-1=0$$$$\left(3\right)-6p^3+12p^2+6pq-6p-2=0$$$$\left(2\right)q=-3p^2+2p+\frac{1}{2}$$$$\left(3\right)-24p^3+24p^2-3p-2=0$$$$p^3-p^2+\frac{1}{8}p=-\frac{1}{12}$$$$$$$$x^4+y^4+z^4=-32(p^3-p^2+\frac{1}{8}p-\frac{3}{64})=\frac{25}{6}$$$$x^5+y^5+z^5=-60(p^3-p^2+\frac{1}{8}p-\frac{1}{60})=6$$$$\mathrm{but}\mathrm{this}\mathrm{method}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{work}\mathrm{for}n⩾6$$

Commented by mr W last updated on 05/Dec/19

$$thanksalotsir!$$

Answered by mind is power last updated on 05/Dec/19

$$\mathrm{x}+\mathrm{y}+\mathrm{z}={\mathrm{s}}_1$$$$\mathrm{xy}+\mathrm{yz}+\mathrm{zx}={\mathrm{s}}_2,$$$$\mathrm{xyz}={\mathrm{s}}_3$$$${\mathrm{p}}_{\mathrm{k}}={\mathrm{x}}^{\mathrm{k}}+{\mathrm{y}}^{\mathrm{k}}+{\mathrm{z}}^{\mathrm{k}}$$$${\mathrm{s}}_{\mathrm{n}}=0,\mathrm{n}⩾4$$$${2\mathrm{s}}_2={\mathrm{s}}_1{\mathrm{p}}_1-{\mathrm{p}}_2=1-2=-1\Rightarrow {\mathrm{s}}_2=-\frac{1}{2}$$$${3\mathrm{s}}_3={\mathrm{s}}_2{\mathrm{p}}_1-{\mathrm{s}}_1{\mathrm{p}}_1+{\mathrm{p}}_2\Rightarrow {3\mathrm{s}}_3=-\frac{1}{2}-1+2\Rightarrow {\mathrm{s}}_3=\frac{1}{6}$$$${4\mathrm{s}}_4={\mathrm{s}}_3{\mathrm{p}}_1-{\mathrm{s}}_2{\mathrm{p}}_2+{\mathrm{s}}_1{\mathrm{p}}_3-{\mathrm{p}}_4=0$$$$\Rightarrow {\mathrm{p}}_4=\frac{1}{6}+1+3=\frac{25}{6}$$$${5\mathrm{s}}_5={\mathrm{s}}_4{\mathrm{p}}_1-{\mathrm{s}}_3{\mathrm{p}}_2+{\mathrm{s}}_2{\mathrm{p}}_3-{\mathrm{s}}_1{\mathrm{p}}_4+{\mathrm{p}}_5=0$$$$\Rightarrow {\mathrm{p}}_5=\frac{2}{6}+\frac{3}{2}+\frac{25}{6}=6$$$${6\mathrm{s}}_6={\mathrm{s}}_3{\mathrm{p}}_3-{\mathrm{s}}_2{\mathrm{p}}_4+{\mathrm{s}}_1{\mathrm{p}}_5-{\mathrm{p}}_6=0$$$${\mathrm{p}}_6=\frac{3}{6}+\frac{25}{12}+6=\frac{103}{12}$$$$\mathrm{x}+\mathrm{y}+\mathrm{z}=1,\mathrm{xy}+\mathrm{xz}+\mathrm{zy}=-\frac{1}{2},$$$$\mathrm{xyz}=\frac{1}{6}$$$$\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{root}\mathrm{of}$$$${\mathrm{S}}^3-{\mathrm{s}}^2-\frac{\mathrm{s}}{2}-\frac{1}{6}=0$$$$\mathrm{cardan},{\mathrm{s}}_1,{\mathrm{s}}_2,{\mathrm{s}}_3\mathrm{roots}$$$${\mathrm{x}}^{\mathrm{n}}+{\mathrm{y}}^{\mathrm{n}}+{\mathrm{z}}^{\mathrm{n}}={\mathrm{s}}_1^{\mathrm{n}}+{\mathrm{s}}_2^{\mathrm{n}}+{\mathrm{s}}_3^{\mathrm{n}}$$$$$$

Commented by mr W last updated on 05/Dec/19

$$thanksalotsir!$$

Commented by mind is power last updated on 05/Dec/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{great}$$

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